Prove $ab \leq (\frac{a+b}{2})^2$ $\forall a,b \in{R^+}$ As the title says I am trying to prove that $ab \leq (\frac{a+b}{2})^2$ $\forall a,b \in{R^+}$.

I tried using induction but the induction step never ends up simplifying the way I want it to. I may be looking at this question in the wrong way or I am just not seeing the "trick" in the induction step. Any help is appreciated! Thanks.

edit: The 2 is inside the parenthesis
Induction is not the right method, since $a,b \in \mathbb R$.

$ab \leq \frac{(a+b)^2}{2} \iff 2ab \le (a+b)^2 \iff 2ab \le a^2+2ab+b^2 \iff a^2+b^2 \ge 0$.
As @dxiv suggests, Result follows from AM-GM inequality, Which says,
If $\{a_i\}$ be a sequence with $a_i \in \mathbb{R}^+ $,

Then, $\frac{1}{n}\sum_{\forall i} a_i \geq \prod_{\forall i} a_i^\frac{1}{n}$ Where. $n$ is the length of the sequence.

Let, $\{a,b\}$ be a sequence, with $a, b \geq 0$; and $n = 2$

Then,

$\frac{a + b}{2} \geq \sqrt{ab} \implies \frac{a + b}{2}^2 \geq {ab}$