JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

# AM GM Inequality Problems

In Mathematics, inequality is a relation that makes a non-equal comparison between mathematical expressions or two numbers. The AMâ€“GM inequality, or inequality of arithmetic and geometric means, states that the arithmetic means of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list.

If every number in the list is the same, only then there is a possibility that two means are equal. In this section, we will practise problems related to AMâ€“GM inequality.

The simplest non-trivial case, i.e., with more than one variable for two non-negative numbers x and y, is the statement that

$$\begin{array}{l}{\frac {x+y}2}\geq {\sqrt {xy}}\end{array}$$

with equality if and only if x = y. This case can be seen from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the elementary case

$$\begin{array}{l}(a \pm b)^2 = a^2 \pm 2ab + b^2\end{array}$$
of the binomial formula:

\begin{array}{l}{\begin{aligned}0&\leq (x-y)^{2}\\&=x^{2}-2xy+y^{2}\\&=x^{2}+2xy+y^{2}-4xy\\&=(x+y)^{2}-4xy.\end{aligned}}\end{array}

Hence, (x + y)2 â‰¥ 4xy, with equality precisely when (x – y)2 = 0, i.e., x = y.

The AMâ€“GM inequality then follows by taking the positive square root of both sides.

For a geometrical interpretation, consider a rectangle with sides of length x and y, hence it has a perimeter 2x + 2y and area xy. Similarly, a square with all sides of length âˆš(xy) has a perimeter of 4âˆš(xy)Â and the same area as the rectangle. The simplest non-trivial case of the AMâ€“GM inequality implies for the perimeters that

$$\begin{array}{l}2x + 2y \geq 4\sqrt{xy}\end{array}$$
and that only the square has the smallest perimeter amongst all rectangles of equal area.

### AM-GM Inequality Relationship

Let a1, a2,…, an be n positive real numbers. The Arithmetic Mean and Geometric Mean are defined as:

$$\begin{array}{l}AM = \frac{a_1 + a_2 + …+ a_n}{n}\end{array}$$
$$\begin{array}{l}GM =\sqrt[n]{a_1 a_2 … a_n}\end{array}$$

The AMâ€“GM inequality states that

AM â‰¥ GM

## Solved Examples on AM-GM Inequality

Example 1: Prove that

$$\begin{array}{l}(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )\ge g\alpha \beta \gamma\end{array}$$

Solution: Applying AM > GM

$$\begin{array}{l}\frac{\alpha +\beta +\gamma }{3}\ge {{\left( \alpha \beta \gamma \right)}^{{}^{1}/{}_{3}}}\,\,\,\,\,………\,(1)\end{array}$$

$$\begin{array}{l}\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{3}\ge {{\left( {{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}} \right)}^{{}^{1}/{}_{3}}}\,\,\,\,\,………\,(2)\end{array}$$

On multiplying equations (1) and (2),

We get

$$\begin{array}{l}\frac{\alpha +\beta +\gamma \left( \alpha \beta +\beta \gamma +\gamma \alpha \right)}{g}\ge {{\left( \alpha \beta \gamma \right)}^{{}^{1}/{}_{3}}}\,{{\left( \alpha \beta \gamma \right)}^{{}^{1}/{}_{3}}}\ge \left( \alpha \beta \gamma \right)\end{array}$$

Therefore,

$$\begin{array}{l} (\alpha +\beta +\gamma )\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)\ge g\alpha \beta \gamma\end{array}$$

Hence proved.

Example 2: Insert 3 GMs between 1 and 16

Solution: Let the terms be 1, g1, g2, g3, 16

$$\begin{array}{l}r={{\left( \frac{16}{1} \right)}^{\frac{1}{3+1}}}\end{array}$$
$$\begin{array}{l}r={{\left( 16 \right)}^{1/4}}=2\end{array}$$

Three geometric means between 1 and 16 are 2, 4, 8.

Example 3: If P, Q, R are positive real numbers such that P + Q + R = g, find the maximum value of P2Q3R4.

Solution: P + Q + R = g

Applying AM > GM

$$\begin{array}{l}2\left( \frac{P}{2} \right)+3\left( \frac{Q}{3} \right)+4\left( \frac{R}{4} \right)=g\end{array}$$
$$\begin{array}{l}\Rightarrow \left( {{\left( \frac{P}{2} \right)}^{2}}.{{\left( \frac{Q}{3} \right)}^{3}}.{{\left( \frac{R}{4} \right)}^{4}} \right)\le \frac{g}{3}\end{array}$$

3g, 22, 33, 44 > P2Q2R4

$$\begin{array}{l}{{3}^{12}}\cdot {{2}^{10}}\ge {{P}^{2}}{{Q}^{3}}{{R}^{4}}\end{array}$$
$$\begin{array}{l}{{6}^{10}}\cdot {{3}^{2}}\ge {{P}^{2}}{{Q}^{3}}{{R}^{4}}\end{array}$$

Example 4: Evaluate 12 + 22 + 32 + 42 + 52.

Solution: Above expression can be written as

$$\begin{array}{l}\sum_{m=1}^5 m^2 \end{array}$$

We know, sum of squares of first five natural numbers = 1/6 {n(n+1)(2n+1)}

Here, n = 5.

Therefore,

$$\begin{array}{l}\sum_{m=1}^5 m^2 \end{array}$$
= 5(5+1)(2.5+1)/6 = 55

Example 5: Prove (a + b + c) (ab + bc + ca) > 9abc.

Solution:

$$\begin{array}{l}\frac{a+b+c}{3}>(abc)^{1/3}\end{array}$$
Â and

$$\begin{array}{l}\frac{ab+bc+ca}{3}> (ab.bc.ca)^{\frac{1}{3}}\end{array}$$
[Using AM > GM]

Multiplying the above results, we get

$$\begin{array}{l}\frac{a+b+c}{3} \times \frac{ab+bc+ca}{3} > (abc)^{1/3}(ab.bc.ca)^{1/3}\end{array}$$

= (a3b3c3)1/3

= abc

â‡’Â (a + b + c)(ab + bc + ca) > 9abc.

Hence proved.

Example 5:Â Consider the functionÂ

$$\begin{array}{l}f(x,y,z)={\frac {x}{y}}+{\sqrt {{\frac {y}{z}}}}+{\sqrt[ {3}]{{\frac {z}{x}}}}\end{array}$$
for all positive real numbersÂ x,Â yÂ andÂ z. Find the minimal value of the function.

Solution:Â

\begin{array}{l}{\begin{aligned}f(x,y,z)&=6\cdot {\frac {{\frac {x}{y}}+{\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}+{\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}+{\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}+{\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}+{\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}}{6}}\\&=6\cdot {\frac {x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}}\end{aligned}}\\ with \ {\displaystyle x_{1}={\frac {x}{y}},\qquad x_{2}=x_{3}={\frac {1}{2}}{\sqrt {\frac {y}{z}}},\qquad x_{4}=x_{5}=x_{6}={\frac {1}{3}}{\sqrt[{3}]{\frac {z}{x}}}.}\end{array}

Applying the AMâ€“GM inequality forÂ nÂ = 6, we get

\begin{array}{l}{\begin{aligned}f(x,y,z)&\geq 6\cdot {\sqrt[ {6}]{{\frac {x}{y}}\cdot {\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}\cdot {\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}\cdot {\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}\cdot {\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}\cdot {\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}}}\\&=6\cdot {\sqrt[ {6}]{{\frac {1}{2\cdot 2\cdot 3\cdot 3\cdot 3}}{\frac {x}{y}}{\frac {y}{z}}{\frac {z}{x}}}}\\&=2^{{2/3}}\cdot 3^{{1/2}}.\end{aligned}}\end{array}

Further, we know that the two sides are equal exactly when all the terms of the mean are equal.

$$\begin{array}{l}f(x,y,z)=2^{{2/3}}\cdot 3^{{1/2}}\text{when}\ {\frac {x}{y}}={\frac {1}{2}}{\sqrt {{\frac {y}{z}}}}={\frac {1}{3}}{\sqrt[ {3}]{{\frac {z}{x}}}}.\end{array}$$

All the points (x, y, z), satisfying these conditions lie on a half-line starting at the origin and are given by,

$$\begin{array}{l}{\displaystyle (x,y,z)={\biggr (}t,{\sqrt[{3}]{2}}{\sqrt {3}}\,t,{\frac {3{\sqrt {3}}}{2}}\,t{\biggr )}\text with \ t \ >0.}\end{array}$$