# Prove that: (a+b)/2 = sqrt( a*b) When a>=0 and b>=0 ?

Oct 1, 2017

$\frac{a + b}{2} \textcolor{red}{\ge} \sqrt{a b} \text{ }$ as shown below

#### Explanation:

Note that:

${\left(a - b\right)}^{2} \ge 0 \text{ }$ for any real values of $a , b$.

Multiplying out, this becomes:

${a}^{2} - 2 a b + {b}^{2} \ge 0$

Add $4 a b$ to both sides to get:

${a}^{2} + 2 a b + {b}^{2} \ge 4 a b$

Factor the left hand side to get:

${\left(a + b\right)}^{2} \ge 4 a b$

Since $a , b \ge 0$ we can take the principal square root of both sides to find:

$a + b \ge 2 \sqrt{a b}$

Divide both sides by $2$ to get:

$\frac{a + b}{2} \ge \sqrt{a b}$

Note that if $a \ne b$ then $\frac{a + b}{2} > \sqrt{a b}$, since then we have ${\left(a - b\right)}^{2} > 0$.