# Inequalities Involving A.M, G.M, H.M

## Inequalities Involving A.M, G.M, H.M

Let A, G and H be Arithmetic, Geometric, and Harmonic means of two positive numbers a and b.

Then, $$A=\frac{a+b}{2}$$, $$G=\sqrt{ab}$$ and $$H=\frac{2ab}{a+b}$$.

These three means possess the following property.

A ≥ G ≥ H.

Proof: We have $$A=\frac{a+b}{2}$$, $$G=\sqrt{ab}$$ and $$H=\frac{2ab}{a+b}$$.

$$A-G=\frac{a+b}{2}-\sqrt{ab}$$,

$$=\frac{a+b-2\sqrt{ab}}{2}$$,

$$=\frac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}{2}$$,

$$=\frac{{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}}{2}\ge 0$$,

A ≥ G … (1)

$$G-H=\sqrt{ab}-\frac{2ab}{a+b}$$,

$$G-H=\frac{-2ab+\left( a+b \right)\sqrt{ab}}{a+b}$$,

$$G-H=\sqrt{ab}\left( \frac{\left( a+b \right)-2\sqrt{ab}}{a+b} \right)$$,

$$=\frac{\sqrt{ab}}{a+b}{{\left( \sqrt{a}-\sqrt{b} \right)}^{2}}\ge 0$$,

G ≥ H … (2)

From Equation (1) and (2), we will get

A ≥ G ≥ H.

Example: Find the minimum value of $${{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}$$.

Solution: AM ≥ GM

$$A.M=\frac{{{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}}{2}$$,

$$G.M=\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}$$,

$$\frac{{{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}}{2}=\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}$$,

$${{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{{{\sin }^{2}}x}}\times {{4}^{{{\cos }^{2}}x}}}$$,

$${{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{{{\sin }^{2}}x+{{\cos }^{2}}x}}}$$,

$${{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\sqrt{{{4}^{\left( 1 \right)}}}$$,

$${{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=2\times 2$$,

$${{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}}=4$$,

$${{\left( {{4}^{{{\sin }^{2}}x}}+{{4}^{{{\cos }^{2}}x}} \right)}_{\min }}=4$$.