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