Inequalities using Arithmetic Mean Geometric Mean


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 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 may ways to prove this Arithmetic Mean and Geometric Mean inequality.

Proofs of Inequalities using Arithmetic Mean Geometric Mean

Proof by square roots

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

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

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

Sum of Reciprocal Fractions

Prove \(\displaystyle \frac{x}{y} + \frac{y}{x} \ge 2\), where \(x\) and \(y\) are positive real numbers.
\( \begin{aligned}
\displaystyle \frac{x}{y} + \frac{y}{x} &\ge 2 \sqrt{\frac{x}{y} \times \frac{y}{x}} &\color{green} \text{arithmetic mean} \ge \text{geometric mean} \\
&= 2 \sqrt{1} \\
&= 2 \\
\therefore \frac{x}{y} + \frac{y}{x} &\ge 2 \\
\end{aligned} \\ \)

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.

Quartic Roots

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

Sum of Reciprocals

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

Sum of Square Reciprocals

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

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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