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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

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

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

\(\begin{array}{l}(x + y)^2 \geq 4xy\end{array} \)
, with equality precisely when
\(\begin{array}{l}[x-y]^{2}=0\end{array} \)
, 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

\(\begin{array}{l}\sqrt{xy}\end{array} \)
has the perimeter
\(\begin{array}{l}4\sqrt{xy}\end{array} \)
and the same area as the rectangle. The simplest non-trivial case of the AM–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 defined as:

AM =

\(\begin{array}{l}\frac{a_1 + a_2 + …+ a_n}{n}\end{array} \)

GM =

\(\begin{array}{l}\sqrt[n]{a_1 a_2 … a_n}\end{array} \)

The AM–GM inequality states that

AM ≥ 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 > G.M

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

 

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

On multiplying equation (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 G.M’s 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 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:

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

\(\begin{array}{l}\frac{ab+bc+ca}{3}\end{array} \)
>
\(\begin{array}{l}(ab.bc.ca)^{\frac{1}{3}}\end{array} \)
.

[Using AM > GM]

Multiplying above results, we get

\(\begin{array}{l}\frac{a+b+c}{3} \times \frac{ab+bc+ca}{3} \end{array} \)
> (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 

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

Solution: 

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

Applying the AM–GM inequality for n = 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

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

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