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
 A: Induction is not the right method, since $a,b \in \mathbb R$.
$ab \leq \frac{(a+b)^2}{2} \iff 2ab \le (a+b)^2 \iff 2ab \le a^2+2ab+b^2 \iff a^2+b^2 \ge 0$.
A: Use polar coordinates
$$a=r\cos \theta,b=r \sin \theta.$$
Your inequality becomes
$$r^2 \cos \theta \sin \theta \leq \frac{r^2}{4} (\cos \theta+\sin \theta)^2=\frac{r^2}{4}(1+2 \cos \theta \sin \theta), $$
which simplifies to
$$\sin (2 \theta) \leq 1.$$
A: It's $a^2+2ab+b^2\geq2ab$ or $a^2+b^2\geq0$
A: First notice
$$(a+b)^2 = a^2 + 2ab + b^2\;.$$
Now divide by 2
$$\frac{(a+b)^2}{2} = \frac{a^2+b^2}{2}+ab\geq ab$$
where the last inequality is because $a^2$ and $b^2$ are positive.
A: working backwards we get $$4ab\le a^2+b^2+2ab$$ and this is equivalent to $$(a-b)^2\geq0$$ which is obviously true.
A: Since $x\in \mathbb{R}$, $x^2\geq 0$,
$(a+b)^2=a^2+b^2+2ab=(a^2+b^2-2ab)+4ab=(a-b)^2+4ab\geq 4ab\space\space\space\blacksquare$
A: 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,
$\frac{a + b}{2} \geq \sqrt{ab} \implies \frac{a + b}{2}^2 \geq {ab}$
