# Prove by Induction AM-GM

Suppose that $a,b \in \mathbb{R}$ are positive.

Prove that:

$$\sqrt{ab} \leq \frac{a + b}{2}$$

Note: This inequality is known as the inequality between arithmetic mean, $\frac{a + b}{2}$, and geometric mean, $\sqrt{ab}$, in short, the am-gm inequality.

Hint: Use the fact that $(x,y)^2 \geq 0$ for all numbers $x$ and $y$.

So far we only learned how to prove by Induction.

My solution:

$\sqrt{ab} \leq \frac{a + b}{2}$ --> can be written as =

$$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$

Base case: $n=1$

$$\frac{x_1}{1} \geq \sqrt[1]{x_1}$$ ---> True $x_1 = x_1$

Inductive step: Assume true for $n=k$

$$\frac{x_1+ \ldots + x_k}{k} \geq \sqrt[k]{x_1 \cdots x_k}$$

Show true for n=k+1

$$\frac{x_1+ \ldots + x_{k+1}}{k+1} \geq \sqrt[k+1]{x_1 \cdots x_{k+1}}$$

Your theorem is only about the $n=2$ case, so I don't see the need to prove a more general statement. Just set $x=\sqrt{a}$, $y=\sqrt{b}$ and follow the hint given: $$0\le(x-y)^2=x^2+y^2-2xy=a+b-2\sqrt{ab},$$ Q.E.D.