AM GM Inequality Problems

In mathematics, inequality is a relation that makes a non-equal comparison between mathematical expressions or two numbers. The AM–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 then only there is a possibility that two means are equal. In this section, we will practice problems related to AM–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

x+y2xy{\frac {x+y}2}\geq {\sqrt {xy}}

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 (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2 of the binomial formula:

0(xy)2=x22xy+y2=x2+2xy+y24xy=(x+y)24xy.{\begin{aligned}0&\leq (x-y)^{2}\\&=x^{2}-2xy+y^{2}\\&=x^{2}+2xy+y^{2}-4xy\\&=(x+y)^{2}-4xy.\end{aligned}}

Hence (x+y)24xy(x + y)^2 \geq 4xy, with equality precisely when [xy]2=0[x-y]^{2}=0, i.e. x = y.

The AM–GM inequality then follows from 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 perimeter 2x + 2y and area xy. Similarly, a square with all sides of length xy\sqrt{xy} has the perimeter 4xy4\sqrt{xy} and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that 2x+2y4xy2x + 2y \geq 4\sqrt{xy} 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 defined as:

AM = a1+a2++ann\frac{a_1 + a_2 + …+ a_n}{n}

GM = a1a2ann\sqrt[n]{a_1 a_2 … a_n}

The AM–GM inequality states that

AM ≥ GM

Solved Examples on AM-GM Inequality

Example 1: Prove that (α+β+γ)(αβ+βγ+γα)gαβγ(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )\ge g\alpha \beta \gamma

Solution: Applying AM > G.M

α+β+γ3(αβγ)1/3(1)\frac{\alpha +\beta +\gamma }{3}\ge {{\left( \alpha \beta \gamma \right)}^{{}^{1}/{}_{3}}}\,\,\,\,\,………\,(1)

 

αβ+βγ+γα3(α2β2γ2)1/3(2)\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{3}\ge {{\left( {{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}} \right)}^{{}^{1}/{}_{3}}}\,\,\,\,\,………\,(2)

On multiplying equation (1) and (2)

We get: –

α+β+γ(αβ+βγ+γα)g(αβγ)1/3(αβγ)1/3(αβγ)\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)

Therefore, (α+β+γ)(αβ+βγ+γα)gαβγ (\alpha +\beta +\gamma )\left( \alpha \beta +\beta \gamma +\gamma \alpha \right)\ge g\alpha \beta \gamma

Hence Proved.

Example 2: Insert 3 G.M’s between 1 and 16

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

r=(161)13+1r={{\left( \frac{16}{1} \right)}^{\frac{1}{3+1}}} r=(16)1/4=2r={{\left( 16 \right)}^{1/4}}=2

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 maximum value of P2Q3R4.

Solution: P + Q + R = g

Applying AM > GM

2(P2)+3(Q3)+4(R4)=g2\left( \frac{P}{2} \right)+3\left( \frac{Q}{3} \right)+4\left( \frac{R}{4} \right)=g ((P2)2.(Q3)3.(R4)4)g3\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}

3g, 22, 33, 44 > P2Q2R4

312210P2Q3R4{{3}^{12}}\cdot {{2}^{10}}\ge {{P}^{2}}{{Q}^{3}}{{R}^{4}} 61032P2Q3R4{{6}^{10}}\cdot {{3}^{2}}\ge {{P}^{2}}{{Q}^{3}}{{R}^{4}}

 

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

Solution: Above expression can be written as m=15m2\sum_{m=1}^5 m^2

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

Here, n = 5.

Therefore, m=15m2\sum_{m=1}^5 m^2 = 5(5+1)(2.5+1)/6 = 55

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

Solution:

a+b+c3\frac{a+b+c}{3} > (abc)1/3 and

ab+bc+ca3\frac{ab+bc+ca}{3} > (ab.bc.ca)13(ab.bc.ca)^{\frac{1}{3}}.

[Using AM > GM]

Multiplying above results, we get

a+b+c3×ab+bc+ca3\frac{a+b+c}{3} \times \frac{ab+bc+ca}{3} > (abc)1/3 (ab.bc.ca)1/3

= (a3b3c3)1/3

= abc

=> (a + b + c)(ab + bc + ca) > 9abc.

Hence Proved.

Example 5: Consider the function f(x,y,z)=xy+yz+zx3f(x,y,z)={\frac {x}{y}}+{\sqrt {{\frac {y}{z}}}}+{\sqrt[ {3}]{{\frac {z}{x}}}}for all positive real numbers xy and z. Find the minimal value of the function.

Solution: 

f(x,y,z)=6xy+12yz+12yz+13zx3+13zx3+13zx36=6x1+x2+x3+x4+x5+x66with x1=xy,x2=x3=12yz,x4=x5=x6=13zx3.{\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}}}.}

Applying the AM–GM inequality for n = 6, we get

f(x,y,z)6xy12yz12yz13zx313zx313zx36=6122333xyyzzx6=22/331/2.{\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}}

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

f(x,y,z)=22/331/2 when xy=12yz=13zx3.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}}}}.

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

(x,y,z)=(t,233t,332t)with t >0.{\displaystyle (x,y,z)={\biggr (}t,{\sqrt[{3}]{2}}{\sqrt {3}}\,t,{\frac {3{\sqrt {3}}}{2}}\,t{\biggr )}\text with \ t \ >0.}