# Inequalities using Arithmetic Mean Geometric Mean

The Arithmetic Mean of $a$ and $b$ is always greater than or equal to the Geometric Mean of $a$ and $b$, for all positive real numbers with equality if and only if $a = b$. This is also called AM-GM (Arithmetic Mean Geometric Mean) inequality.
$\require{color}$
\begin{aligned} \frac{a + b}{2} \ge \sqrt{ab} \text{ or } a + b \ge 2 \sqrt{ab} \end{aligned}
There are many ways to prove this Arithmetic Mean and Geometric Mean inequality.

## Proofs of Inequalities using Arithmetic Mean Geometric Mean

### Proof by square roots

\begin{aligned} \displaystyle \left(\sqrt{a}-\sqrt{b}\right)^2 &\ge 0 &\color{green} \text{square of anything is positive or zero} \\ a-2 \sqrt{ab} + b &\ge 0 &\color{green} \text{expand} \\ \therefore a + b &\ge 2 \sqrt{ab} &\color{green} \text{got it!} \end{aligned}

### Proof by subtraction

\begin{aligned} (a + b)^2-4ab &= a^2 + 2ab + b^2-4ab \\ &= a^2-2ab + b^2 \\ &= (a-b)^2 \ge 0 \\ (a + b)^2-4ab &\ge 0 \\ (a + b)^2 &\ge 4ab \\ \therefore a + b &\ge 2\sqrt{ab} \end{aligned}

### Proof by subtraction and square roots

\begin{aligned} a + b-2 \sqrt{ab} &= \sqrt{a}^2-2 \sqrt{a} \sqrt{b} + \sqrt{b}^2 \\ &= \left(\sqrt{a}-\sqrt{b}\right)^2 \ge 0 \\ a + b-2 \sqrt{ab} &\ge 0 \\ \therefore a + b &\ge 2 \sqrt{ab} \end{aligned}

### Sum of Reciprocal Fractions

Prove $\displaystyle \frac{x}{y} + \frac{y}{x} \ge 2$, where $x$ and $y$ are positive real numbers.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \frac{x}{y} + \frac{y}{x} &\ge 2 \sqrt{\frac{x}{y} \times \frac{y}{x}} &\color{green} \text{arithmetic mean} \ge \text{geometric mean} \\ &= 2 \sqrt{1} \\ &= 2 \\ \therefore \frac{x}{y} + \frac{y}{x} &\ge 2 \end{aligned}

### Product of sum and reciprocals

Prove $\displaystyle (x + y)\Big(\frac{1}{x} +\frac{1}{y}\Big) \ge 4$ , where $x$ and $y$ are positive real numbers.

#### Method 1

\begin{aligned} \displaystyle (x + y)\Big(\frac{1}{x} +\frac{1}{y}\Big) &\ge 2 \sqrt{x \times y} \times 2 \sqrt{\frac{1}{x} \times \frac{1}{y}} &\color{green} \text{arithmetic mean} \ge \text{geometric mean} \\ &= 4 \sqrt{\frac{xy}{xy}} \\ &= 4 \\ \therefore (x + y)\Big(\frac{1}{x} +\frac{1}{y}\Big) &\ge 4 \end{aligned}

#### Method 2

\begin{aligned} \displaystyle (x + y)\Big(\frac{1}{x} +\frac{1}{y}\Big) &= 1 + \frac{x}{y} + \frac{y}{x} + 1 \\ &\ge 1 + 2 \sqrt{\frac{x}{y} \times \frac{y}{x}} + 1 &\color{green} \text{arithmetic mean} \ge \text{geometric mean} \\ &= 4 \\ \therefore (x + y)\Big(\frac{1}{x} +\frac{1}{y}\Big) &\ge 4 \end{aligned}

### Product of sum and reciprocals in three terms

Prove $\displaystyle (x + y + z)\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \Big) \ge 9$ , where $x, y$ and $z$ are positive real numbers.

\begin{aligned} \displaystyle (x + y + z)\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \Big) &= 1 + \frac{x}{y} + \frac{x}{z} + \frac{y}{x} + 1 + \frac{y}{z} + \frac{z}{x} + \frac{z}{y} + 1 &\color{green} \text{expand} \\ &= 3 + \Big(\frac{x}{y} + \frac{y}{x} \Big) + \Big(\frac{x}{z} + \frac{z}{x} \Big) + \Big(\frac{y}{z} + \frac{z}{y} \Big) &\color{green} \text{group each pairs} \\ &\ge 3 + 2 \sqrt{\frac{x}{y} \times \frac{y}{x}} + 2 \sqrt{\frac{x}{z} \times \frac{z}{x}} + 2 \sqrt{\frac{z}{y} \times \frac{y}{z}} &\color{green} \frac{x}{y} + \frac{y}{x} \ge 2 \sqrt{\frac{x}{y} \times \frac{y}{x}} \\ &= 3 + 2 + 2 + 2 \\ &= 9 \\ \therefore (x + y + z)\Big(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \Big) &\ge 9 \end{aligned}

### Quartic Roots

Prove $\displaystyle a + b + x + y \ge 4 \sqrt[4]{abxy}$, where $a,b,x,y \in \Re^{+}$.

\begin{aligned} \displaystyle \require{AMSstyle} \require{color} a + b + x + y &\ge 2 \sqrt{ab} + 2 \sqrt{xy} &\color{green} a + b \ge 2 \sqrt{ab} \text{ and } x + y \ge 2 \sqrt{xy} \\ &\ge 2 \sqrt{2 \sqrt{ab} \times 2 \sqrt{xy}} &\color{green} \text{arithmetic mean} \ge \text{geometric mean} \\ &= 4 \sqrt{\sqrt{ab} \times \sqrt{xy}} \\ &= 4 \sqrt{\sqrt{abxy}} \\ &= 4 \sqrt[4]{abxy} \\ \therefore a + b + x + y &\ge 4 \sqrt[4]{abxy} \end{aligned}

### Sum of Reciprocals

Prove $\displaystyle \frac{1}{x} + \frac{1}{y} \ge \frac{4}{x+y}$, where $x,y \in \Re^{+}$.

\begin{aligned} \displaystyle \frac{1}{x} + \frac{1}{y} &\ge 2 \sqrt{\frac{1}{x} \times \frac{1}{y}} \\ &= \frac{2}{\sqrt{xy}} &\color{green} (1) \\ x + y &\ge 2 \sqrt{xy} \\ \frac{1}{2\sqrt{xy}} &\ge \frac{1}{x + y} &\color{green} \text{upside down both sides} \\ \frac{2}{\sqrt{xy}} &\ge \frac{4}{x + y} &\color{green} (2) \\ \frac{1}{x} + \frac{1}{y} &\ge \frac{2}{\sqrt{xy}} \ge \frac{4}{x+y} &\color{green} \text{by }(1) \text{ and } (2) \\ \therefore \frac{1}{x} + \frac{1}{y} &\ge \frac{4}{x+y} \end{aligned}

### Sum of Square Reciprocals

Prove $\displaystyle\frac{1}{x^2} + \frac{1}{y^2} \ge \frac{8}{(x + y)^2}$, where $x,y \in \Re^{+}$.

\begin{aligned} \require{AMSsymbols} \require{color} \frac{1}{x^2} + \frac{1}{y^2} &\ge 2 \sqrt{\frac{1}{x^2} \times \frac{1}{y^2}} \\ &= \frac{2}{xy} \\ \frac{1}{x^2} + \frac{1}{y^2} &\ge \frac{2}{xy} &\color{green} (1) \\ x + y &\ge 2 \sqrt{xy} \\ (x + y)^2 &\ge 4xy \\ \frac{1}{4xy} &\ge \frac{1}{(x + y)^2} \\ \frac{2}{xy} &\ge \frac{8}{(x + y)^2} &\color{green} (2) \\ \frac{1}{x^2} + \frac{1}{y^2} &\ge \frac{2}{xy} \ge \frac{8}{(x + y)^2} &\color{green} \text{by }(1) \text{ and } (2) \\ \therefore \frac{1}{x^2} + \frac{1}{y^2} &\ge \frac{8}{(x + y)^2} \end{aligned}

### Sum of Squares

Prove $x^2 + y^2 \ge \sqrt{2}$, if $x^4 + y^4 = 1$.

\begin{aligned} x^4 + y^4 &\ge 2 \sqrt{x^4 y^4} \\ &= 2 x^2 x^2 \\ 1 &\ge 2x^2 y^2 &\color{green} (1) \\ (x^2 + y^2)^2 &= x^4 + 2x^2 y^2 + y^4 \\ &= 1 + 2 x^2 y^2 \\ &\ge 1 + 1 &\color{green} \text{by }(1)\\ &= 2 \\ (x^2 + y^2)^2 &\ge 2 \\ \therefore x^2 + y^2 &\ge \sqrt{2} \end{aligned}

### Sum of Exponentials

Prove $\displaystyle e^a + e^b \ge 2 e^{\frac{a+b}{2}}$.

\begin{aligned} \displaystyle e^a + e^b &\ge 2 \sqrt{e^a \times e^b} \\ &= 2 \sqrt{e^{a+b}} \\ &= 2 e^{\frac{a+b}{2}} \\ \therefore e^a + e^b &\ge 2 e^{\frac{a+b}{2}} \end{aligned}

### Minimum Value of Exponentials

Find the minimum value of $e^{2x} + e^x + e^{-x} + e^{-2x}$.

\begin{aligned} e^{2x} + e^x + e^{-x} + e^{-2x} &\ge 2 e^{\frac{2x-2x}{2}} + 2 e^{\frac{x-x}{2}} \\ &= 2 e^0 + 2 e^0 \\ &= 2 + 2 \\ &= 4 \\ e^{2x} + e^x + e^{-x} + e^{-2x} &\ge 4 \\ \text{Therefore the minimum value is 4.} \end{aligned}