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Last modified on May 17th, 2024

AM-GM inequality, also known as inequality of arithmetic or geometric means, states that the arithmetic mean of any group of positive real numbers is greater than its geometric mean, and they are equal if and only if the chosen numbers are the same.

Mathematically, it is written as

If â€˜x_{1},â€™ â€˜x_{2},â€™ â€¦, â€˜x_{n}â€™ (â‰¥ 0) are the real numbers, then the inequality

${\dfrac{x_{1}+x_{2}+\ldots +x_{n}}{n}\geq \sqrt[n] {x_{1}x_{2}\ldots x_{n}}}$

Here, the equality holds if and only if x_{1} = x_{2} = â€¦ = x_{n}

Thus,

${\dfrac{\sum ^{n}_{i=1}x_{i}}{n}\geq \sqrt[n] {\prod ^{n}_{i=1}x_{i}}}$

Cauchyâ€™s induction is a type of mathematical induction (also known as forward-backward induction) in which we prove the result for 2 variables, 2n variables, and then for (n – 1) variables.

**Base Step (For 2 Variables):**

Using the properties of perfect squares, we get

(x – y)^{2} â‰¥ 0, with the equality if and only if x – y = 0 â‡’ x = y

Since x and y are positive real numbers, we have

x^{2} – 2xy + y^{2} â‰¥ 0

â‡’ x^{2} – 2xy + 4xy + y^{2} â‰¥ 0 + 4xy

â‡’ x^{2} + 2xy + y^{2} â‰¥ 4xy

â‡’ (x + y)^{2} â‰¥ 4xy

â‡’ ${\dfrac{\left( x+y\right) ^{2}}{4}\geq xy}$

â‡’ ${\dfrac{x+y}{2}\geq \sqrt{xy}}$, where the equality holds if and only if x = y

Thus, the base step is proved.

**Forward Step (For 2n Variables):**

Let us assume the inequality holds true for â€˜nâ€™ variables. Now, we prove that the inequality holds for â€˜2n.â€™

Let â€˜x_{1},â€™ â€˜x_{2},â€™ â€¦, â€˜x_{2n}â€™ be the group of positive real numbers, where â€˜x_{1},â€™ â€˜x_{2},â€™ â€¦, â€˜x_{n}â€™ and â€˜x_{n + 1},â€™ â€˜x_{n + 2},â€™ â€¦, â€˜x_{2n}â€™ are two sub-groups of â€˜nâ€™ variables.

Thus, ${\dfrac{x_{1}+x_{2}+\ldots +x_{n}}{n}\geq \sqrt[n] {x_{1}x_{2}\ldots x_{n}}}$ and

${\dfrac{x_{n+1}+x_{n+2}+\ldots +x_{2n}}{n}\geq \sqrt[n] {x_{n+1}x_{n+2}\ldots x_{2n}}}$

On adding the above inequalities, we get

${\dfrac{x_{1}+x_{2}+\ldots +x_{2n}}{n}\geq \sqrt[n] {x_{1}x_{2}\ldots x_{n}}+\sqrt[n] {x_{n+1}x_{n+2}\ldots x_{2n}}}$

Now, dividing both sides by 2, we get

${\dfrac{x_{1}+x_{2}+\ldots +x_{2n}}{2n}\geq \dfrac{\sqrt[n] {x_{1}x_{2}\ldots x_{n}}+\sqrt[n] {x_{n+1}x_{n+2}\ldots x_{2n}}}{2}}$ â€¦..(i)

Also, from AM-GM inequality for two variables ${\sqrt[n] {x_{1}x_{2}\ldots x_{n}}}$ and ${\sqrt[n] {x_{n+1}x_{n+2}\ldots x_{2n}}}$, we get

${\dfrac{\sqrt[n] {x_{1}x_{2}\ldots x_{n}}+\sqrt[n] {x_{n+1}x_{n+2}\ldots x_{2n}}}{2n}\geq \sqrt[2n] {x_{1}x_{2}\ldots x_{2n}}}$ â€¦..(ii)

By combining (i) and (ii), we get

${\dfrac{x_{1}+x_{2}+\ldots +x_{2n}}{2n}\geq \sqrt[2n] {x_{1}x_{2}\ldots x_{2n}}}$, which is the AM-GM inequality in â€˜2nâ€™ variables with the equality if and only if

${\sqrt[n] {x_{1}x_{2}\ldots x_{n}}}$ = ${\sqrt[n] {x_{n+1}x_{n+2}\ldots x_{2n}}}$

Thus, x_{1} = x_{2} = â€¦ = x_{2n}, the equality condition for AM-GM inequality in â€˜2nâ€™ variables.

Thus, the forward step is proved.

**Backward Step (For (n – 1) Variables):**

Let us assume the inequality holds true for â€˜nâ€™ variables. Now, we prove that the inequality holds for (n – 1) variables.

Let ${x_{n}=\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}}$

By substituting the value of â€˜x_{n}â€™ in AM-GM inequality for â€˜nâ€™ variables, we get

Since the inequality is true for â€˜nâ€™ variables, equality holds if and only if x_{1} = x_{2} = â€¦ = x_{n – 1} = ${\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}}$, which is only possible if x_{1} = x_{2} = â€¦ = x_{n – 1}

Calculating the L.H.S of (iii), we get

${\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}+\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}}{n}}$

= ${\dfrac{nx_{1}+nx_{2}+\ldots +nx_{n-1}}{n\left( n-1\right) }}$

= ${\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{\left( n-1\right) }}$

Now, putting this into (iii), we get

${\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{\left( n-1\right) }\geq \sqrt[n] {x_{1}x_{2}\ldots x_{n-1}\left( \dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{\left( n-1\right) }\right) }}$

Raising both sides to the n^{th} power, we get

${\left( \dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}\right) ^{n}\geq x_{1}x_{2}\ldots x_{n-1}\left( \dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}\right)}$

On dividing both sides by ${\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}}$, we get

${\left( \dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}\right) ^{n-1}\geq x_{1}x_{2}\ldots x_{n-1}}$

By taking the (n – 1)^{th} root, we get

${\dfrac{x_{1}+x_{2}+\ldots +x_{n-1}}{n-1}\geq \sqrt[n-1] {x_{1}x_{2}\ldots x_{n-1}}}$, which is the inequality for (n – 1) variables.

Thus, the backward step is proved, and the AM-GM inequality is proved by Cauchyâ€™s induction.

Let us define â€˜nâ€™ numbers of sequences ${\left\{ r_{i,j}\right\} _{i=1}^{n}}$ such that ${r_{i,j}=\sqrt[n] {x_{i}}}$, where 1 â‰¤ i, j â‰¤ n

By Rearrangement inequality, we get

${\sum _{i}x_{i}=\sum _{i}\prod _{j}r_{i,j}\geq \sum _{i}\prod _{j}r_{i+j,j}=n\prod _{i}\sqrt[n] {x_{i}}}$

â‡’ ${\sum _{i}x_{i}\geq n\prod _{i}\sqrt[n] {x_{i}}}$, where the equality holds if and only if all â€˜r_{i, j}â€™ and all â€˜x_{i}â€™ are equal

Now, dividing both sides by â€˜n,â€™ we get

${\dfrac{\sum ^{n}_{i=1}x_{i}}{n}\geq \sqrt[n] {\prod ^{n}_{i=1}x_{i}}}$

Thus, the AM-GM inequality is proved.

Moreover, the AM-GM inequality is further generalized into several other inequalities.

It states that if â€˜Ï‰_{1},â€™ â€˜Ï‰_{2},â€™ â€¦, â€˜Ï‰_{n}â€™ (â‰¥ 0) are a group of weights such that Ï‰_{1} + Ï‰_{2} + â€¦ + Ï‰_{n} = Ï‰, then

${\dfrac{\omega _{1}x_{1}+\omega _{2}x_{2}+\ldots +\omega _{n}x_{n}}{\omega }\geq \sqrt[\omega ] {x_{1}^{\omega _{1}}x_{2}^{\omega _{2}}\ldots x_{n}^{\omega _{n}}}}$, where the equality hold if and only if x_{1} = x_{2} = â€¦ = x_{n}

â‡’ ${\dfrac{\sum ^{n}_{i=1}\omega _{i}x_{i}}{\omega }\geq \sqrt[\omega ] {\prod ^{n}_{i=1}x_{i}^{w_{i}}}}$

When Ï‰_{1} = Ï‰_{2} = â€¦ = Ï‰_{n} = ${\dfrac{1}{n}}$, the weighted AM-GM inequality reduces to the AM-GM inequality.

Since the second-order derivative of the logarithmic function x â†’ ln(x) is less than or equal to 0, the function is strictly concave, which satisfies the condition of Jensenâ€™s inequality.

By Jensenâ€™s inequality,

${\ln \left( \dfrac{\omega _{1}x_{1}+\omega _{2}x_{2}+\ldots +\omega _{n}x_{n}}{\omega }\right) \geq \dfrac{\omega _{1}}{\omega }\ln x_{1}+\ldots +\dfrac{\omega _{n}}{\omega }\ln x_{n}}$

â‡’ ${\ln \left( \dfrac{\omega _{1}x_{1}+\omega _{2}x_{2}+\ldots +\omega _{n}x_{n}}{\omega }\right) \geq \ln \left( \sqrt[\omega ] {x_{1}^{\omega _{1}}x_{2}^{\omega _{2}}\ldots x_{n}^{\omega _{n}}}\right)}$

â‡’ ${\dfrac{\sum ^{n}_{i=1}\omega _{i}x_{i}}{\omega }\geq \sqrt[\omega ] {\prod ^{n}_{i=1}x_{i}^{w_{i}}}}$, where Ï‰_{1} + Ï‰_{2} + â€¦ + Ï‰_{n} = Ï‰

Thus, the weighted AM-GM inequality is proved.

The mean inequality chain, also known as the **RMS-AM-GM-HM inequality**, states that

**E.g.1. If x, y, z are positive real numbers and x + y + z = 1, then prove that x ^{x}y^{y}z^{z} + x^{y}y^{z}z^{x} + x^{z}y^{x}z^{y} â‰¤ 1.Â **

Solution:

By AM-GM inequality, we get

${\dfrac{x^{2}+y^{2}+z^{2}}{x+y+z}\geq \left( x^{x}y^{y}z^{z}\right) ^{\dfrac{1}{x+y+z}}}$

â‡’ ${x^{2}+y^{2}+z^{2}\geq x^{x}y^{y}z^{z}}$ (since x + y + z = 1) â€¦..(i)

Again, by AM-GM inequality, we get

${\dfrac{xy+yz+zx}{x+y+z}\geq \left( x^{y}y^{z}z^{x}\right) ^{\dfrac{1}{x+y+z}}}$

â‡’ ${xy+yz+zx\geq x^{y}y^{z}z^{x}}$ (since x + y + z = 1) â€¦..(ii)

Again, by AM-GM inequality, we get

${\dfrac{xz+yx+zy}{x+y+z}\geq \left( x^{z}y^{x}z^{y}\right) ^{\dfrac{1}{x+y+z}}}$

â‡’ ${xz+yx+zy\geq x^{z}y^{x}z^{y}}$ (since x + y + z = 1) â€¦..(iii)

On adding both sides of (i), (ii), and (iii), we get

(x + y + z)^{2} â‰¥ x^{x}y^{y}z^{z} + x^{y}y^{z}z^{x} + x^{z}y^{x}z^{y}Â

â‡’ 1 â‰¥ x^{x}y^{y}z^{z} + x^{y}y^{z}z^{x} + x^{z}y^{x}z^{y}Â

â‡’ x^{x}y^{y}z^{z} + x^{y}y^{z}z^{x} + x^{z}y^{x}z^{y} â‰¤ 1

**Prove that (m ^{2}n + n^{2}p + p^{2}m)(mn^{2} + np^{2} + pm^{2}) â‰¥ 9m^{2}n^{2}p^{2}Â **

Solution:

By AM-GM inequality, we get

${\dfrac{m^{2}n+n^{2}p+p^{2}m}{3}\geq \left( m^{3}n^{3}p^{3}\right) ^{\dfrac{1}{3}}}$Â

â‡’ ${m^{2}n+n^{2}p+p^{2}m\geq 3mnp}$ â€¦..(i)

Again, by AM-GM inequality, we get

${\dfrac{mn^{2}+np^{2}+pm^{2}}{3}\geq \left( m^{3}n^{3}p^{3}\right) ^{\dfrac{1}{3}}}$Â

â‡’ ${mn^{2}+np^{2}+pm^{2}\geq 3mnp}$ â€¦..(ii)

On multiplying (i) and (ii), we get(m^{2}n + n^{2}p + p^{2}m)(mn^{2} + np^{2} + pm^{2}) â‰¥ 9m^{2}n^{2}p^{2}

Last modified on May 17th, 2024