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
$$\frac{x_1+ \ldots + x_{k+1}}{k+1} \geq \sqrt[k+1]{x_1 \cdots x_{k+1}}$$
I am stuck at this step. Please help Thank you very much.
