<s>[INST] Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that 
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
 [/INST] Expand both sides and we have:
$$\frac{x^2 + y^2}{2} \ge \frac{x^2 + 2xy + y^2}{4}$$
$$\iff x^2 + y^2 \ge {2xy}$$
$$\iff (x-y)^2 \ge 0$$
Which is well-know fact. 
Also there are lot of other ways to prove it, here's "fancy" one using Cauchy-Scwarz inequality:
$$(1 + 1)(x^2 + y^2) \ge (x + y)^2$$
Multiply both sides by $\frac 14$:
$$\frac 12 (x^2 + y^2) \ge \left(\frac 12 (x+y)\right)^2$$
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