# Maxima Minima : AM-GM Concept

Maxima Minima : AM-GM Concept
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## AM-GM-HM Concept

A wide variety of questions are being framed in this area. There is no thumb rule to solve all these problems. Many a time, simple observation and hit and trial methods of putting the values work. However, to get the results quickly and accurately, there are specific categories of questions which require knowledge of few concepts. Let us discuss them one by one.

For any positive real numbers a1, a2, a3, ….., an. The following relation holds:

R.M.S. ≥ A.M. ≥ G.M. ≥ H.M. i.e

$\sqrt {\frac{{a_1^2 + a_2^2 + a_3^2 + …. + a_n^2}}{n}}$$\ge$ $\frac{{{a_1} + {a_2} + {a_3} + …. + {a_n}}}{n}$$\ge$ $\sqrt[n]{{a{}_1{a_2}{a_3}…..{a_n}}}$$\ge$ $\frac{n}{{\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + .. + \frac{1}{{{a_n}}}}}$

Here: RMS = Root means square, A.M. = Arithmetic Mean, G.M. = Geometric Mean, H.M. = Harmonic Mean.

The equality holds true if all the values are equal.

Note:

• If the sum of variables is constant, the product of the variables will be maximum when all of them are equal.
• If the product of variables is constant, the sum is minimum when all of them are equal

### AM-GM Concept Solved Example

Example: For a positive real number a, b, and c, if a + b + c = 15, find the maximum value of abc?

Solution:

AM ≥ GM

$\frac{{a + b + c}}{3} \ge \sqrt[3]{{a \times b \times c}}$

Or $\frac{{15}}{3} \ge \sqrt[3]{{a \times b \times c}}$

Or, $abc \le {5^3}$. Hence the maximum value of abc = 125, when a= b = c = 5.

Example: For positive real number x, y, and z,  If xyz = 64, find the minimum value of x+y+z?

Solution:

Clearly, the sum is minimum if all the variables are equal, i.e. x = y = z = 4. Hence the minimum value of x + y + z = 4+4+4 = 12.

Example: Let a, b and c be nonnegative integers such that a + b+ c = 15. What is the maximum value of a.b.c + a.b + b.c + c.a?

Solution:

(a+1) (b+1)(c+1) = a.b.c + a.b + b.c + c.a + a + b + c + 1

= a.b.c + a.b + b.c + c.a + 16

Applying AM-GM

$\frac{{{\rm{(a + 1) }} + {\rm{(b + 1)}} + {\rm{(c + 1) }}}}{3} \ge \sqrt[3]{{{\rm{(a + 1) (b + 1)(c + 1) }}}}$

Thus (a+1) (b+1)(c+1) is maximum at 216, which occurs when a = b =c = 5. Therefore a.b.c + a.b + b.c + c.a = 200.

### Some Special cases

In many questions, we have to manipulate the problems so that the direct formulas can be applied. Let see some examples to learn the approaches.

### Case I:

Example: for positive real number x and y, if 2x + 3y = 15, find the maximum value of x2y.

Solution:

Observe that, if we directly apply AM-GM relation then, we will get the term xy and NOT x2y. For the maximum value of x2y, there should be three terms among which two terms are containing variable x and one term containing variable y. So that, when we multiply the terms, we get x2y.

We have to do all this without distorting the equation given in the problems.

Now, 2x + 3y = 15 can be written as x + x + 3y = 15. Here assuming both xs to be different terms resulting in equation with three terms. i.e. x, x and 3y. Now we can apply the relation AM ≥ GM.

$\frac{{x + x + 3y}}{3} \ge \sqrt[3]{{x \times x \times 3y}}$

Or $\frac{{15}}{3} \ge \sqrt[3]{{{x^2} \times 3y}}$

Or $\sqrt[3]{{{x^2} \times 3y}} \le 5$

Cubing both side we get

${x^2}\left( {3y} \right) \le 125$ or x2y $\le$125/3. Hence the maximum value is 125/3

### Case II:

Example: Let x, y and z be distinct positive integers such that the product x.y.z = 2001. What is the largest possible value of the sum x+y+z?

Solution:

We know that, if all the variables are equal then the product is maximum, and for the given product, the sum is minimum if all the variables are equal. But here we need to find the maximum value of the sum. Therefore we should keep the values of all the variables as far as possible. i.e.

$xyz = 2001 = 1 \times 3 \times 667$

Taking x = 1, y = 3 and z = 667, we get the maximum sum = 1+ 3+667 = 671.

### Case III:

Example: If a1, a2, a3 and a4 are positive integers with sum = 16. Find the minimum value of $\left( {\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + \frac{1}{{{a_4}}}} \right)$?

Solution:

AM $\ge$ HM

$\frac{{{a_1} + {a_2} + {a_3} + {a_4}}}{4} \ge \frac{4}{{\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + \frac{1}{{{a_4}}}}}$

Or $\left( {\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + \frac{1}{{{a_4}}}} \right)$$\left( {{a_1} + {a_2} + {a_3} + {a_4}} \right)$≥42s

Hence minimum value of $\left( {\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + \frac{1}{{{a_4}}}} \right)$ = 16/16 = 1

### Case IV:

Example: If three positive real numbers x, y and z are in A.P. such that xyz=4, then what will be the minimum value of y.

Solution:

If x, y and z are in AP, then 2y = x + z.

Also, AM $\ge$ GM

$\frac{{x + y + z}}{3} \ge \sqrt[3]{{xyz}}$

Or $\frac{{3y}}{3} \ge \sqrt[3]{4}$

Or $y \ge \sqrt[3]{4} = {2^{\frac{2}{3}}}$

### Case V:

Example: Minimize the expression (x+y)(y+z), where x, y and z are positive real numbers satisfying xyz(x + y + z) = 1.

Solution:

(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 AM, GM inequality gives $(x + y)(y + z) = xz + \frac{1}{{xz}} \ge 2\sqrt {xz \times \frac{1}{{zx}} = } 2$

## Cauchy Inequality

For any real numbers a1, …….., an and b1,……..,bn

${\left( {\sum\limits_{i = 1}^n {{a_i}{b_i}} } \right)^2} \le \left( {\sum\limits_{i = 1}^n {{a^2}_i} } \right)\left( {\sum\limits_{i = 1}^n {{b^2}_i} } \right)$ with equality when there exist constants µ, β not both zero such that for all 1≤i≤n,  µai = β bi

Example: Find the largest and smallest value of 2x+3y+6z for values of x, y, z satisfying x2 + y2 + z2=1.

Solution:

By Cauchy inequality with n = 3

(22 + 32 + 62)(x2 + y2 + z2) ≥(2x+3y+6z)2 with equality if $\frac{x}{2} = \frac{y}{3} = \frac{z}{6}$

Therefore, (2x+3y+6z)2 ≤ 49.

Or -7≤ (2x+3y+6z) ≤ 7. Hence the minimum and maximum values are -7 and 7 respectively.

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