merged document

Using the Arithmetic Mean-Geometric Mean Inequality in Problem Solving.

http://jwilson.coe.uga.edu/EMT725/AMGM/SSMA.Bnghm.html[11/9/12 3:34:14 PM]

Using the Arithmetic Mean-Geometric Mean Inequality in Problem Solvingby

JIM WILSON

A Presentation to the Annual Meeting of School Mathematics and Science Association, Birmingham, November 8, 2012, was prepared using some parts of thispaper.

PDF version

The Arithmetic Mean-Geometric Mean Inequality (AM-GM Inquality) is a fundamentalrelationship in mathematics. It is a useful tool for problems solving and buildingrelationships with other mathematics. It should find more use in school mathematics thancurrently. It what follows I present an introduction to the theorem, some background andgeneralizations, alternative demonstrations of the proof, and examples of problems that canbe explored by using the AM-GM Inquality. I will rely heavily on a collection of problemsand essays on my web site:

Http://jwilson.coe.uga.edu

and the sub-directory for my mathematics problem solving course at the University ofGeorgia.

http://jwilson.coe.uga.edu/EMT725/AMGM/Using%20the%20AM-GM%20in%20Problem%20Solving.pdf

http://jwilson.coe.uga.edu/

http://jwilson.coe.uga.edu/EMT725/EMT725.html



I will concentrate on the theorem for two positive numbers in my examples, but Imention the generalizations below and occasionally use the case for three positive numbers.

Arithmetic Mean - Geometric Mean Inequality For real positive numbers a and b, the AM-GM Inequality for two numbers is:



Clearly, the AM-GM Inequality can be generalized for n positive numbers. The link posesthe problem of generalizing the proof following the lines of argument advanced by Courantand Robbins (1942). That is,

http://jwilson.coe.uga.edu/EMT725/AMGM/Generalize.html



The AM-GM Inequality can also be generalized to its inclusion in relation to othermeans such as the Harmonic Mean (HM) or the Root Mean Square (RMS -- sometimescalled the Quadratic Mean). In particular, for two positive numbers a and b,

http://jwilson.coe.uga.edu/EMT725/HM/HM.html

http://jwilson.coe.uga.edu/EMT725/Isos.Trpzd/isos.trpzd.html



Geometric demonstration of the RMS-AM-GM-HM Inequality

At a more advanced level (perhaps more fundamental?) all of these means areinstances of Power Means where the power parameter p takes on different values for thedifferent means. These may also be called Generalized Means.

What is the Value of this Theorem? An Example. The AM-GM for two positive numbers can be a useful tool in examining someoptimization problems. For example, it is well known that for rectangles with a fixedperimeter, the maximum area is given by a square having that perimeter.

http://jwilson.coe.uga.edu/EMT725/Isos.Trpzd/LinesInTrpzd.gsp

http://mathworld.wolfram.com/PowerMean.html



Perhaps some insight is given by the graph at the right. The fixed perimeteris

2a + 2b = 10.

The red graph represents the area function of the graph as either of a or bvaries from 0 to 5. Let b represent the length of one side and 5 - b thelength of the other.



Area = ab = b(5 - b) for 0 < b < 5

When the AM-GM Inequality is applied we get Area ²

Points on the blue curve, Area = 6.25 are always greater than points on the red curve (That is,the area of the rectangle is always less that 6.25) and the blue curve (a horizontal linerepresenting a constant) is tangent to the red curve if and only if b = (5 - b), i.e., b =2.5.

Setting this up with the usual function notation, let the Perimeter of the rectangle equal 10and let one side be x. The the other side is 5 - x.

The utility of the AM-GM Inequality is that the replacement function after the application ofthe AM-GM inequality is a constant line tangent to the previous function. The area is alwaysless than the constant AND it is equal to that constant when a = b. In our example for P =10, a = b = 2.5 when the rectangle is a square.

From my web page, here is a problem posed for students:

http://jwilson.coe.uga.edu/EMT725/AMGM/AMGMrec/index.html

http://jwilson.coe.uga.edu/EMT725/MaxRec/MaxRec.html



The second problem, where ab is a constant rather than a + b, is a nice contrast to the firstand it also follows quickly from the AM-GM Inequality:



Alternative Proofs and Demonstrations

of the

Arithmetic Mean-Geometric Mean Inequality In this section, I will limit the exploration to the simplest case: The arithmetic mean andgeometric mean of two positive numbers. In my Problem Solving course I pose AS ANEXPLORATION that the student find a least 5 demonstrations or proofs. Seeing multipleapproaches to this relation can help with understanding it, seeing its importance, andfinding it useful as a problem solving tool.

Some algebra

http://jwilson.coe.uga.edu/EMT725/AMGM/AMGM.3.html



A Geometric Demonstration




Another Geometric Suggestion

Construct a semicircle with a diameter a +b. The radius will be the Arithmetic Mean of a and b. Construct a perpendicular to thediameter from common endpoint of thesegments of length a and b. From theintersection of this perpendicular with thesemicircle construct the red segment. Thissegment will always have a length less thanor equal to the radius of the circle and it will be equal only if a = b.




This example is closely related to the well known geometric theorem thatthe altitude of a right triangle from the 90 degree vertex to thehypothenuse will be the geometric mean of the two segments cut off onthe hypotenuse.

Another Geometry Example




Another Algebra Demonstration

This demonstration begins with the identity:

Since we have with equality if and only if a= b.

Another Geometry Example

Given two tangent circles of radii a and b. Construct a common externaltangent to the two circl3s and draw radii of each circle to the common tangent. Find the length indicatec by ? along the common tanget in terms of a and b.

Solution: by constructing a segment parallel to the one under study, a righttriangle with legs of length ? and a-b, and hypotenuse of length a+b. thesegment along the common tangent has length twice the geometric mean of aand b.

More Geometry

Consider a circle with chords AB and CD. Let theCD pass through the midpoint M of AB.

Let AM = MB = z and let CM = x, MD = y. Thelengths are all positive values.






By an elementary theorem of geometry theproducts of the parts of the chords are equal.

We know that x + y ³ 2z with equality only if M is the midpoint of CD.

Algebra

Something Different





Solving Problems with the AM-GM Inequality

Cost of Fencing a Field

Problem: A farmer wants to fence in 60,000 square feet of land in a rectangular plotalong a straight highway. The fence he plans to use along the highway costs $2 perfoot, while the fence for the other three sides costs $1 per foot. How much of each typeof fence will he have to buy in order to keep expenses to a minimum? What is theminimum expense?

Solution: (This is a typical calculus problem but no calculus is needed!)

http://jwilson.coe.uga.edu/emt725/Fence/Fence.html



Minimum of

The problem is to find minimum values for this function in therange or 0 < x < π. This one requires some change in the formof the equation in order to apply the AM-GM Inequality. First,however, it is helpful to see a graph and have some sense of theequation. The graph at the right suggests that there may be twominimum values in the range 0 < x < π.

http://jwilson.coe.uga.edu/emt725/SineMin/SineMin.html





Construct a Square with Same Area as a Given Rectangle

The problem is to construct a squarewith straightedge and compass that isthe same area as a given rectangle.

The area will be ab and so the lengthof the side of the square is the geometricmean.

http://jwilson.coe.uga.edu/EMT725/SquareToRectangle/Rectangle.html



Consider:

The red segment is the geometric mean of a and b.

Maximum Area of a Sector of a Circle With Fixed Perimeter

Understanding the problem. A sector of a circle has a perimeter made up of two radiiand an arc of the circle connecting the endpoints of the two radii. Compare the areasof three sectors -- each with P = 100 -- central angles of 45 degrees, 90 degrees, and180 degrees. These are sectors that are an eighth circle, quarter circle, and semicircle.As the angles increas, the radii become shorter because more of the fixed perimeter isin the arc. Note these angles are

When these three areas are computed we get

http://jwilson.coe.uga.edu/EMT725/Sector/sector.html



Clearly, as the angle increases from 45 to 90 to 180 the area increases and thendecreases. Somewhere there is a maximum area. Where?

Sector as Fraction of a Circle

Some students (and teachers) have an aversion to working with radianmeasure. For them the approach might be as follows:

Let k be the fraction of a circle represented by the sector. 0 < k< 1. The Perimeter is a constant and so we can represent it as twicethe radius plus the fraction of the circumference.

The area of the sector in the fraction notation can be formed as either a function ofr or a function of k, depending on a subsitution of r or k from the perimeterequation. Either one can be used with the AM-GM Inequality to reach closure on theproblem:



Looking back at the example when P = 100, the maximum area would bewhen r = 25 and therefore the arc length is 50. The angle is a little lessthan 120 degrees.

Sector in Radian Measure



The parallel of this solution with the one where the square was themaximum area for rectangles with fixed perimeter is worth noting.

Inequalities Problems

http://jwilson.coe.uga.edu/EMT725/Inequalities/Inequalities.html



Comparison of altitude and median in a right triangle

Construct segments from the endpointsof the diameter to complete a righttriangle.

http://jwilson.coe.uga.edu/EMT725/Bob2/MedAlt/MedAltRtTri.html



Maximum and Minimum of

Rewrite

For x > 0, using the arithmetic mean-geometric mean inequality,

Therefore the value of the function is always less than or equal to .5 and it isequal to .5 only when x = 1.

For x < 0, a similar argument leads to finding the minimum of the function at x= -1. Since the Arithmetic Mean -- Geometric Mean Inequality holds only for

http://jwilson.coe.uga.edu/EMT725/min/MinMax.html

http://jwilson.coe.uga.edu/EMT725/min/MinMax.html



postitive values, when x < 0 we have to apply the inequality to - x and - 1/x. Weknow

Keep in mind this is for x< 0 so -x and -1/x are postive. Multiplying each side ofthe inequality by -1 gives

and equality occurs when x = -1. Therefore the value of the function is alwaysmore than or equal to -0.5 and it is equal to -0.5 only when x = -1.See Graphs

Maximum of f(x) = (1-x)(1+x)(1+x)

We have three factors in the function and we want to know when the function reachesa maximum in the interval [0,1].

In order to take advantage of the AM-GM Inequality, we need the sum of these factorsto equal a constant. That is we need -2x rather than -x. We can get that by thefollowing

http://jwilson.coe.uga.edu/EMT725/AMGM/MinMax/index.html

http://jwilson.coe.uga.edu/EMT725/MaxCubic/MaxCubic.html



The function reaches this maximum value for x in [0,1] if and only if

2(1 - x) = 1 + x 2 - 2x = 1 + x 1 = 3x

That is,

Minimum Surface Area of a Can of Fixed Volume

In packaging a product in a can the shape of right circular cylinder,various factors such as tradition and supposed customer preferencesmay enter into decisions about what shape (e.g. short and fat vs. talland skinny) can might be used for a fixed volume. Note, for example,all 12 oz. soda cans have the same shape -- a height of about 5 inches

http://jwilson.coe.uga.edu/EMT725/MinSurf/Minimum.Surface.Area.html



and a radius of about 1.25 inches. Why?

What if the decison was based on minimizing the material used tomake the can? This would mean that for a fixed volume V the shapeof the can (e.g. the radius and the height) would be determined by theminimum surface area for the can. What is the relationship between the radius andthe height in order to minimize the surface area for a fixed volume?

What would be the shape of a 12 ounce soda can that minimizes the amount ofaluminum in the can?

The Volume is fixed.

The Surface area is a function of r and h.

What is the minimum S?



( Note that we have use the AM-GM Inequality for threepositive numbers)



Maximum area of a Pen, fixed length of fencing, with Partitions Parallel to a Side

Suppose you have 100 feet of fencing. You want a rectangular pen with a partitionparallel to a side. The 100 feet of fencing must enclose the four sides and thepartition. What is the shape of the pen and what is the maximum area?

Show that the dimensions of the maximum area pen is 25 ft. by 16.67 ft. with maximarea of approximately 417.7 sq. ft.

Prove the Maximum Area of a Triangle with Fixed Perimeter is Equilateral

http://jwilson.coe.uga.edu/EMT725/AMGM/PenPartition.gsp

http://jwilson.coe.uga.edu/EMT725/AMGM/EquilateralProof.gsp



Minimum Distance from (0,0) to

Find a point on the graph of that is closest to the origin.

(follow the link to the discussion)

Maximum Area of a Triangle with Sides of 9, 40x, and 41xGiven a triangle with one side oflength 9 units and the ratio of theother two sides is 40/41.

Find the Maximum Area.

(use link for discussion)

The Box ProblemUse the Arithmetic Mean-Geometric Mean Inequality to find the maximum volume ofa box made from a 25 by 25 square sheet of cardboard by removing a small squarefrom each corner and folding up the sides to form a lidless box.

http://jwilson.coe.uga.edu/EMT725/Optimize/MinDist.html

http://jwilson.coe.uga.edu/EMT725/Optimize/MinDist.html

http://jwilson.coe.uga.edu/EMT725/Optimize/MaxTriangle/MaxTri.html

http://jwilson.coe.uga.edu/EMT725/Box/Box.Sol.html



Theorem: In the Product of n positive numbers is equal to 1 then the sum if the numbersis greater than or equal to n.



Clearly, the proof of 3 implies the first two.

Inequalities and ‘Maximum-Minimum’ Problems

Henry Liu, 26 February 2007

There are many olympiad level problems in mathematics which belong to areasthat are not covered well at all at schools. Three major examples are geometry,number theory, and functional equations. Such areas must be learned outside classtime if one wants to be successful at solving olympiad style problems.

This set of notes considers another one of these areas: inequalities. This areais not covered well at schools mainly because there is a large variety of methodsinvolved. One subarea of inequalities involves problems of the following form:

“Maximise or minimise some expression, subject to some constraint.”

We shall discuss some basic facts about inequalities, and then discuss these so-called ‘max-min’ problems.

1. Well-known inequalities

1.1 The basics

We begin with some obvious rules.

Theorem 1 For real numbers a, b, c, d, we have the following.

(a) If a > b, then a+ c > b+ c; if a ≥ b, then a+ c ≥ b+ c.

(b) If a > b and c > d, then a+ c > b+ d; if a ≥ b and c > d, then a+ c > b+ d;if a ≥ b and c ≥ d, then a+ c ≥ b+ d.

(c) If a > b, then ac > bc if c > 0, and ac < bc if c < 0; if a ≥ b, then ac ≥ bc ifc > 0, and ac ≤ bc if c < 0.

(d) Let a, b, c, d > 0. If a > b and c > d, then ac > bd; if a ≥ b and c > d, thenac > bd; if a ≥ b and c ≥ d, then ac ≥ bd;

(e) a2 ≥ 0 always holds, with equality if and only if a = 0.

(f) Let a, b > 0. If a2 > b2, then a > b; if a2 ≥ b2, then a ≥ b.

1.2 Proving inequalities: some well-known techniques

How do we prove an inequality at olympiad level? We will discuss many differenttechniques. But firstly, note that such a problem usually has two parts:

(a) Prove the inequality.

(b) If possible, state when we have equality.

1

It is a common mistake to ignore part (b).

1.2.1 Squares are non-negative

A complicated looking inequality can be proved if we can show that it is equiv-alent to an inequality of the form ‘LHS ≥ 0’, where ‘LHS’ is a sum of squares.

Example 1. Prove that x4 − 7x2 + 4x + 20 ≥ 0 for every real x. When do wehave equality?

We have x4−7x2+4x+20 = (x4−8x2+16)+(x2+4x+4) = (x2−4)2+(x+2)2 ≥ 0.For equality, we need both x2− 4 = 0 and x+ 2 = 0. This is only possible when

x = −2.Example 2. Let a, b ≥ 0. Prove that a+b

2≥√ab.

Since a, b ≥ 0,√a and

√b are well-defined. We have 0 ≤ (

√a −√b)2 =

a + b − 2√ab, so a + b ≥ 2

√ab, so a+b

2≥√ab. Equality holds if and only if√

a−√b = 0, ie: when a = b since a, b ≥ 0.

Example 3. Prove that x2 + y2 + z2 ≥ xy + yz + zx for all real x, y, z.We have 0 ≤ (x− y)2 + (y− z)2 + (z− x)2 = 2x2 + 2y2 + 2z2− 2xy− 2yz− 2zx.

So 2x2 + 2y2 + 2z2 ≥ 2xy + 2yz + 2zx, and we have the result by dividing by 2.Equality holds if and only if x− y = y − z = z − x = 0, ie: when x = y = z.

1.2.2 The QM-AM-GM-HM inequalities

We now describe a chain of inequalities which will be extremely useful.

Theorem 2 (The QM-AM-GM-HM inequalities) Let x1, . . . , xn ≥ 0. We de-fine their quadratic mean to be

QM =

√x2

1 + · · ·+ x2n

n.

Define their arithmetic mean to be

AM =x1 + · · ·+ xn

n.

Define their geometric mean to be

GM = n√x1 · · ·xn.

For x1, . . . , xn > 0, define their harmonic mean to be

HM =n

1x1

+ · · ·+ 1xn

.

Then, if x1, . . . , xn > 0, we have

QM ≥ AM ≥ GM ≥ HM.

If we remove ‘HM’ from above, we may have x1, . . . , xn ≥ 0 instead. ie: Forx1, . . . , xn ≥ 0, we have QM ≥ AM ≥ GM.

We have QM = AM = GM = HM if and only if x1 = · · · = xn.

2

Taking any two of these means gives us a useful inequality. For example, the‘AM-GM inequality’ says: For x1, . . . , xn ≥ 0, we have

x1 + · · ·+ xnn

≥ n√x1 · · ·xn,

with equality if and only if x1 = · · · = xn. Note that Example 2 above is the AM-GM inequality for two terms.

The ‘AM-HM inequality’ says: For x1, . . . , xn > 0, we have

x1 + · · ·+ xnn

≥ n1x1

+ · · ·+ 1xn

,

with equality if and only if x1 = · · · = xn.And so on.Example 4. Let a, b, c, x, y, z > 0. Prove that(

a

x+b

y+c

z

)(x

a+y

b+z

c

)≥ 9.

Applying the AM-HM inequality, we have

1

3

(a

x+b

y+c

z

)≥ 3

xa

+ yb

+ zc

.

This easily rearranges to the required inequality.Equality holds if and only if a, b, c, x, y, z satisfy a

x= b

y= c

z.

1.2.3 Cauchy-Schwarz inequality

This is another well-known inequality which is extremely useful.

Theorem 3 (Cauchy-Schwarz inequality) Let x1, . . . , xn, y1, . . . , yn be real num-bers. Then

(x1y1 + · · ·+ xnyn)2 ≤ (x21 + · · ·+ x2

n)(y21 + · · ·+ y2

n).

Equality holds if and only if either there exists a real number k such that x1 =ky1, x2 = ky2, . . . , xn = kyn, or there exists a real number ` such that y1 = `x1,y2 = `x2, . . . , yn = `xn.

Note: Be careful! The statements ‘there exists a real number k such that x1 =ky1, x2 = ky2, . . . , xn = kyn’ and ‘there exists a real number ` such that y1 = `x1,y2 = `x2, . . . , yn = `xn’ are not quite equivalent. We will have a problem if either allxi = 0 or all yi = 0. For example, if x1 = · · · = xn = 0 and y1, . . . , yn are arbitrary,not all zero, then the first statement above is true with k = 0, but the second isfalse!

3

Remarkably, almost all literature about the Cauchy-Schwarz inequality are nottoo careful about this, merely just stating one of these statements as the conditionfor equality!

Example 5. Let x, y, z > 0. Prove that

(x2 + y2 + z2)1/2 ≥ 1

13(3x+ 4y + 12z).

Applying Cauchy-Schwarz, we have

(32 + 42 + 122)(x2 + y2 + z2) ≥ (3x+ 4y + 12z)2,

x2 + y2 + z2 ≥ 1

169(3x+ 4y + 12z)2.

Since both sides are positive, the result follows by taking square roots.Equality holds if and only if x = 3k, y = 4k and z = 12k for some real number

k > 0.

1.2.4 Jensen’s inequality

This is yet another very useful inequality. To state this inequality, we must firstdefine a property of functions.

A function f(x) defined on an interval I is said to be strictly convex on I if forevery a, b ∈ I with a < b, and for every λ with 0 < λ < 1, we have

f((1− λ)a+ λb) < (1− λ)f(a) + λf(b).

Graphically, this just says that, no matter where we choose a and b within I, thechord joining (a, f(a)) and (b, f(b)) lies strictly above the graph of the function inthe interval (a, b).

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................a b(1−λ)a+λb

f((1−λ)a+λb)

(1−λ)f(a)+λf(b)

(a,f(a))

(b,f(b))

y=f(x)

....

For example, f(x) = x2 is strictly convex on (−∞,∞). g(x) = sin x is strictlyconvex on [π, 2π].

We say that f(x), defined on interval I, is strictly concave on I if −f(x) is strictlyconvex on I. Graphically, for all a, b ∈ I with a < b, the chord joining (a, f(a)) and(b, f(b)) lies strictly below the graph of the function in the interval (a, b).

We can now state Jensen’s inequality.

4

Theorem 4 (Jensen’s inequality) Let f(x) be a function defined on an intervalI, and is strictly convex on I. Then, for every x1, . . . , xn ∈ I, we have

1

n(f(x1) + · · ·+ f(xn)) ≥ f

(1

n(x1 + · · ·+ xn)

).

Equality holds if and only if x1 = · · · = xn.A similar statement holds when we replace ‘convex’ by ‘concave’, and reverse the

inequality.

Example 6. For w, x, y, z > 0, prove that 16(w3+x3+y3+z3) ≥ (w+x+y+z)3.Since f(x) = x3 is strictly convex on (0,∞), by Jensen’s inequality, we have

1

4(w3 + x3 + y3 + z3) ≥

(w + x+ y + z

4

)3

.

This rearranges to the required inequality. Equality holds if and only if w = x =y = z.

Unfortunately, it is not always easy to decide whether a function is strictlyconvex/concave on a certain interval. For example, where is the function f(x) =x4 + 4x3 − 18x2 + 5x strictly convex/concave? One way to determine this would beto use calculus, which we will come to next.

2. Calculus

2.1 Calculus or no calculus?

Despite the fact that a lot of references suggest that the knowledge of calculusis never required in olympiad style problems, calculus is still a useful tool to have.Of course, it has advantages and disadvantages. A big advantage is that it is quitepowerful and sometimes it offers a ‘cheap way out’; namely there may be elegantsolutions to a certain problem, but calculus offers an easy, ugly solution (but stilla solution!). Another is that calculus offers a way to determine where a function isstrictly convex/concave. A big disadvantage is that calculus only locates local min-ima/local maxima/points of inflexion. A detailed analysis of the function concernedis needed to show whether a local minimum/local maximum is also indeed global.Moreover, we may need to compute the second derivative to determine the natureof stationary values, and this is not always pleasant.

Nevertheless, if the function concerned is simple enough and a detailed analysisis carried out, then calculus can offer solutions.

There is the following important principle in calculus as well.

Theorem 5 Suppose f : I → R is twice differentiable on the interval I. Supposethat c ∈ I is the only critical point of f(x). Then, f(c) is the global minimum off(x) if f ′′(c) > 0, and f(c) is the global maximum of f(x) if f ′′(c) < 0.

5

Example 7. For x > 0, minimise 16x

+ 81x3/2.At first sight, this problem is crying out for a calculus solution. Let f(x) =

16x−1 + 81x3/2. We have f ′(x) = −16x−2 + 2432x1/2. Solving f ′(x) = 0 gives x = 4

9.

We find that f(49) = 60. Also, f ′′(x) = 32x−3 + 243

4x−1/2. We see that f ′′(x) > 0 for

all x > 0, so in particular, f ′′(49) > 0. Hence by Theorem 5, f(4

9) = 60 is the global

minimum of f(x), and 60 is the required minimum. We have equality if and only ifx = 4

9.

We may also argue without Theorem 5, by considering the graph of f(x). Ob-serve that f(x) → ∞ when x → 0 and when x → ∞. Indeed f(x) has the y-axisas an asymptote, and it behaves like 81x3/2 for large x. The graph of f(x) looks like:

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.....

................

................

y

60

0 49

x

f(x)= 16x

+81x3/2

.

We could, in fact, use the AM-GM inequality to solve this problem: apply AM-GM on 16

3x, 16

3x, 16

3x, 81

2x3/2, 81

2x3/2. This does not seem as obvious as the calculus

approach.As mentioned, another thing that calculus can do is to determine where a function

is strictly convex/concave. We have the following result.

Theorem 6 Suppose f : I → R is twice differentiable on the interval I. Then, f(x)is strictly convex on I if f ′′(x) > 0 for all x ∈ I, and f(x) is strictly concave on Iif f ′′(x) < 0 for all x ∈ I.

Example 8. Determine where the function f(x) = x4 + 4x3 − 18x2 + 5x isstrictly convex and strictly concave.

We find easily that f ′′(x) = 12x2 + 24x − 36 = 12(x + 3)(x − 1). We havef ′′(x) > 0 when x < −3 or when x > 1, and f ′′(x) < 0 when −3 < x < 1. Soby Theorem 6, f(x) = x4 + 4x3 − 18x2 + 5x is strictly convex on (−∞,−3) and on(1,∞), and strictly concave on (−3, 1).

In a nutshell, use calculus only when the function is sufficiently simple, and avoidcalculus if the function is too complicated.

3. ‘Maximum-Minimum Problems’

3.1 How do we tackle them?

Problems of the type

6

“Maximise and/or minimise (expression A), subject to (constraint B)”

can be difficult to solve. The difficulty usually seems to be to decide how to incor-porate constraint B into the problem of maximising and/or minimising expressionA. This gets even more difficult if, for example, expression A involves more thantwo variables and constraint B is only a single equation relating the variables: theremay be little or no hope if we try to ‘eliminate one variable’, and one would have tothink of another approach.

The techniques that we have already discussed more or less cover many of thestrategies on how to solve such problems. One just needs to be a little clever indeciding what technique(s) to use.

Finally, once the maximum and/or minimum has/have been decided, one mustthen show that it/they may be attained by giving an example.

Example 9. Minimise the expression (x+y)(y+z), where x, y and z are positivereal numbers satisfying xyz(x+ y + z) = 1.

We have

(x+ y)(y + z) = xy + xz + y2 + yz = xz + y(x+ y + z) = xz +1

xz,

where we have used the condition xyz(x + y + z) = 1 to get the last equality.Applying the AM-GM inequality gives

(x+ y)(y + z) = xz +1

xz≥ 2

√xz · 1

xz= 2.

To claim that ‘2’ is indeed the required minimum value, we must give an exampleof x, y, z where this minimum can be attained, and satisfying the constraint. In theinequality above, we have equality when xz = 1

xz, so xz = 1. To make things simple,

choose x = z = 1. Then y must satisfy y(2 +y) = 1; solving this gives y = −1±√

2.Since y > 0, we take the root with the positive sign. It is then easy to check that(x+ y)(y + z) = 2 when x = z = 1 and y = −1 +

√2.

Example 10. Maximise the expression 7x + 2y + 8z, where x, y and z arepositive real numbers satisfying 4x2 + y2 + 16z2 = 1.

We try to cleverly apply the Cauchy-Schwarz inequality. We have

(7x+ 2y + 8z)2 ≤ ((2x)2 + y2 + (4z)2)

((7

2

)2

+ 22 + 22

)=

81

4,

where we have used the condition 4x2 + y2 + 16z2 = 1 to get the last equality. Wesee that 7x+ 2y + 8z ≤ 9

2.

Again we must show that ‘92’ can be attained by some values of x, y, z satisfying

the given constraint. For equality to hold in Cauchy-Schwarz, we have 2x = 72k,

y = 2k, 4z = 2k for some k > 0. Substituting these into 4x2 + y2 + 16z2 = 1 gives494k2 + 4k2 + 4k2 = 1, giving k = 2

9. This leads to x = 7

18, y = 4

9and z = 1

9. It is

7

then easy to check that with these values of x, y, z, we do have 4x2 + y2 + 16z2 = 1and 7x+ 2y + 8z = 9

2.

4. Problems

Here are some inequalities and ‘maximum-minimum’ problems.

1. Prove that, if x is a real number and x 6= 0, then

x8 − x5 − 1

x+

1

x4≥ 0.

2. Let x and y be real numbers. Prove that (x3 + y3)(x5 + y5) ≤ 2(x8 + y8).

3. Minimise 2x2 +y2 +z2, where x, y and z are real numbers satisfying x+y+z =10.

4. Let x and y be non-zero real numbers satisfying x2+y2 = 4. Find the minimumvalue of

x4 +1

x4+ y4 +

1

y4.

5. (a) Maximise the expression x2y − y2x when 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.(b) Maximise the expression x2y+y2z+z2x−y2x−z2y−x2z when 0 ≤ x ≤ 1,0 ≤ y ≤ 1 and 0 ≤ z ≤ 1.

6. Determine the smallest value of x2 + 5y2 + 8z2, where x, y and z are realnumbers satisfying yz + zx + xy = −1. Does x2 + 5y2 + 8z2 have a greatestvalue subject to the same constraint?

7. Let x, y and z be positive real numbers satisfying

1

3≤ xy + yz + zx ≤ 3.

Determine the range of values of (a) xyz and (b) x+ y + z.

8. Let x, y and z be positive real numbers satisfying xyz = 32. Find the minimumvalue of x2 + 4xy + 4y2 + 2z2.

9. Let x and y be non-negative real numbers satisfying x + y = 2. Show thatx2y2(x2 + y2) ≤ 2.

10. Let x, y and z be real numbers. Prove that

x6 + y6 + z6 + 3x2y2z2 ≥ 2(x3y3 + y3z3 + z3x3).

8

11. Let x, y and z be positive real numbers satisfying x + y + z = 1. Find theminimum value of (

1 +1

x

)(1 +

1

y

)(1 +

1

z

).

12. Let x, y and z be non-negative real numbers satisfying x + y + z = 1. Provethat

7(xy + yz + zx) ≤ 2 + 9xyz.

13. Let x and y be real numbers such that 7x2 + 3xy + 3y2 = 1. Show that theleast positive value of x2+y2

yis 1

2.

14. Let x, y and z be non-negative real numbers satisfying x + y + z = 1. Provethat

x2y + y2z + z2x ≤ 4

27.

15. Let a, b, c be positive real numbers satisfying a+ b+ c = 1. Prove that

a2 + b2 + c2 + 2√

3abc ≤ 1.

16. Let a, b, c be real numbers such that a, b, c ≥ −34

and a+ b+ c = 1. Prove that

a

a2 + 1+

b

b2 + 1+

c

c2 + 1≤ 9

10.

17. Let a, b, c, d be positive real numbers satisfying ab + bc + cd + da = 1. Provethat

a3

b+ c+ d+

b3

c+ d+ a+

c3

d+ a+ b+

d3

a+ b+ c≥ 1

3.

18. Let a, b, c be positive real numbers satisfying abc = 1. Prove that

ab

a5 + b5 + ab+

bc

b5 + c5 + bc+

ca

c5 + a5 + ca≤ 1.

9

This is a guest article by Mr.Dipak Singh.

Today we’ll see how to find the maximum value (greatest value ) or the minimum value (least value)

of a trigonometric function without using differentiation. Take a pen and note-book, keep doing the

steps while reading this article.

First Remember following identities:

Trig-Identities 1. sin2 θ + cos2 θ = 1

2. 1+ cot2 θ = cosec2 θ

3. 1+ tan2 θ = sec2 θ

how did we get these formulas? Already explained, click me

Min-Max table Min value Max value Can be written as

sin θ, sin 2θ, sin 9θ …. sin nθ -1 +1 -1 ≤ Sin nθ ≤ 1

cos θ, cos 4θ , cos 7θ … cos nθ -1 ≤ Cos nθ ≤ 1

sin2 θ , sin2 4θ , sin2 9θ …sin2 nθ 0 +1 Can be written as

0 ≤ Sin2 nθ ≤ 1

cos2 θ , cos2 3θ , cos2 8θ … cos2 nθ 0 ≤ Cos2 nθ ≤ 1

Sin θ Cos θ -1/2 +1/2 -1/2 ≤ Sin θ Cos θ ≤ ½

observe that in case of sin2θ and cos2θ, the minimum value if 0 and not (-1). Why does this happen?

because (-1)2=+1

Negative Signs inside out Sin (- θ) = - Sin (θ)

Cos (-θ) = Cos (θ)

Ratta-fication formulas 1. a sin θ ± b cos θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }

2. a sin θ ± b sin θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }

3. a cos θ ± b cos θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }

4. Min. value of (sin θ cos θ)n = (½)n

http://mrunal.org/2013/04/trigonometry-type4-questions-based-on-trig-and-algebra-combo.html#109

The AM GM Logic Let A ,B are any two numbers then,

Arithmetic Mean (AM)= (A + B) / 2 and

Geometric Mean (GM) = √ (A.B)

Hence, A.M ≥ G.M ( We can check it by putting any values of A and B )

Consider the following statement “ My age is greater than or equal to 25 years . ”

What could you conclude about my age from this statement ?

Answer : My age can be anywhere between 25 to infinity … means it can be 25 , , 50 ,99, 786 or

1000 years etc… but it can not be 24 or 19 or Sweet 16 . Infact it can not be less than 25, strictly.

Means, We can confidently say that my age is not less 25 years. Or in other words my minimum

age is 25 years.

Showing numerically, if Age ≥ 25 years ( minimum age = 25 )

Similarly, If I say x ≥ 56 ( minimum value of x = 56 )

If, y ≥ 77 ( minimum value of y = 77 )

If, x + y ≥ 133 ( minimum value of x + y = 133 )

If, sin θ ≥ - 1 ( minimum value of Sin θ = -1 )

If, tan θ + cot θ ≥ 2 (minimum value of tan θ + cot θ = 2 ) ]]

Sometimes, we come across a special case of trigonometric identities like to find min. value of sin θ +

cosec θ or tan θ + cot θ or cos2 θ + sec2 θ etc. These identities have one thing in common i.e., the

first trigonometric term is opposite of the second term or vice-versa ( tan θ = 1/ cot θ , sin θ = 1/

cosec θ , cos2 θ = 1/ sec2 θ ).

These type of problems can be easily tackled by using the concept of

A.M ≥ G .M

Meaning, Arithmetic mean is always greater than or equal to geometric mean. For example:

Find minimum value of 4 tan2θ + 9 cot2θ (they’ll not ask maximum value as it is not defined. )

We know that tan2θ = 1/ cot2θ , hence applying A.M ≥ G.M logic, we get

A.M of given equation = (4 tan2θ + 9 cot2θ) / 2 …. (1)

G.M of given equation = √ (4 tan2θ . 9 cot2θ )

= √ 4 * 9 # ( tan2θ and cot2θ inverse of each other, so tan x cot =1)

= √ 36 = 6 …. (2)

Now, we know that A.M ≥ G. M

From equations (1) and (2) above we get,

=> (4 tan2 θ + 9 cot2θ) / 2 ≥ 6

Multiplying both sides by 2

=> 4 tan2 θ + 9 cot2 θ ≥ 12 ( minimum value of tan2 θ + cot2 θ is 12 )

Deriving a common conclusion: Consider equation a cos2 θ + b sec2 θ ( find minimum value)

As, A.M ≥ G.M

(a cos2 θ + b sec2 θ / 2 ) ≥ √ (a cos2 θ . b sec2 θ)

a cos2 θ + b sec2 θ ≥ 2 √ (ab) ( minimum value 2 √ab )

So, we can use 2 √ab directly in these kind of problems.

Summary: While using A.M ≥ G.M logic :

Term should be like a T1 + b T2 ; where T1 = 1 / T2

Positive sign in between terms is mandatory. (otherwise how would you calculate mean ? )

Directly apply 2√ab .

Rearrange/Break terms if necessary -> priority should be given to direct use of identities -> find

terms eligible for A.M ≥ G.M logic -> if any, apply -> convert remaining identities, if any, to sine

and cosines -> finally put known max., min. values.

Extra facts: The reciprocal of 0 is + ∞ and vice-versa.

The reciprocal of 1 is 1 and -1 is -1.

If a function has a maximum value its opposite has a minimum value.

A function and its reciprocal have same sign.

Keeping these tools (not exhaustive) in mind we can easily find Maximum or Minimum values easily.

SSC CGL 2012 Tier II Question What is The minimum value of sin2 θ + cos2 θ + sec2 θ + cosec2 θ + tan2 θ + cot2 θ

A. 1

B. 3

C. 5

D. 7

Solution: We know that sin2 θ + cos2 θ = 1 (identitiy#1)

Therefore,

(sin2 θ + cos2 θ) + sec2 θ + cosec2 θ + tan2 θ + cot2 θ

= (1) + sec2 θ + cosec2 θ + tan2 θ + cot2 θ

Using A.M ≥ G.M logic for tan2 θ + cot2 θ we get ,

= 1 + 2 + sec2 θ + cosec2 θ

changing into sin and cos values

( Because we know maximum and minimum values of Sin θ, Cos θ :P and by using simple identities

we can convert all trigonometric functions into equation with Sine and Cosine.)

= 1 + 2 + (1/ cos2 θ) + (1/ sin2 θ)

solving taking L.C.M

= 1 + 2 + (sin2 θ + cos2 θ)/( sin2 θ . cos2 θ)…..eq1

but we already know two things

sin2 θ + cos2 θ=1 (trig identity #1)

Min. value of (sin θ cos θ)n = (½)n (Ratta-fication formula #4)

Apply them into eq1, and we get

= 1 + 2 + (sin2 θ + cos2 θ)/( sin2 θ . cos2 θ)

= 1 + 2 + (1/1/4) = 1+2+4

= 7 (correct answer D)

The least value of 2 sin2 θ + 3 cos2 θ (CGL2012T1) A. 1

B. 2

C. 3

D. 5

We can solve this question via two approaches

Approach #1 Break the equation and use identity no. 1

= 2 sin2 θ + 2 cos2 θ + cos2 θ

=2(sin2 θ + cos2 θ) + cos2 θ ; (but sin2 θ + cos2 θ=1)

= 2 + cos2 θ ;(but as per min-max table, the minimum value of cos2 θ=0)

= 2 + 0 = 2 (correct answer B)

Approach #2 convert equation into one identity ,either sin or cos

first convert it into a sin equation :

= 2 sin2 θ + 3 (1- sin2 θ) ;(because sin2 θ + cos2 θ=1=>cos2 θ=1- sin2 θ)

= 2 sin2 θ + 3 – 3 sin2 θ

= 3 - sin2 θ

= 3 – ( 1) = 2 (but Min. value of sin2 θ is 0 …confusing ???? )

As sin2 θ is preceded by a negative sign therefore we have to take max. value of sin2 θ in order to get

minimum value .

Converting into a cos equation :

= 2 sin2 θ + 3 cos2 θ

= 2 (1- cos2 θ) + 3 cos2 θ

= 2 – 2 cos2 θ + 3 cos2 θ

= 2 + cos2 θ

= 2 + 0 = 2 ( correct answer B )

The maximum value of Sin x + cos x is A. √2

B. 1/ √2

C. 1

D. 2

Applying Ratta-fication formulae No.1

a sin θ ± b cos θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }

in the given question, we’ve to find the max value of

Sin x + cos x

= + √ (12+ 12 )

= √2 ( correct answer A )

The maximum value of 3 Sin x – 4 Cos x is A. -1

B. 5

C. 7

D. 9

Solution:

Applying Ratta-fication formulae No.1

a sin θ ± b cos θ = ±√ (a2 + b2 ) { for min. use - , for max. use + }

in the given question, we’ve to find the max value of

3 Sin x – 4 Cos x

= + √ (32+ 42 )

= √25

= 5 ( correct answer B )

Min Max values of sin 4x + 5 are (A) 2, 6

(B) 4, 5

(C) -4, -5

(D) 4, 6

Solution:

We know that, -1 ≤ Sin nx ≤ 1

= -1 ≤ Sin 4x ≤ 1

Adding 5 throughout, 4 ≤ Sin 4x +5 ≤ 6

Therefore, the minimum value is 4 and maximum value is 6 ( correct answer D )

Minimum and maximum value of Sin Sin x is A. Do not exist

B. -1, 1

C. Sin -1 , Sin +1

D. - Sin 1 , Sin 1

We know that, -1 ≤ Sin nx ≤ 1

= Sin (-1) ≤ Sin x ≤ Sin (1)

= - Sin 1 ≤ Sin x ≤ Sin 1 ; [Sin(-θ) is same as – Sin θ ]

Therefore, Minimum value is –Sin 1 and maximum is Sin 1 ( correct answer D)

The key to success is Practice! Practice! Practice!

Drop your problems in the comment box.

For more articles on trigonometry and aptitude, visit mrunal.org/aptitude

http://www.mrunal.org/aptitude

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