<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] let investigate $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}=\frac{x^2+y^2+2xy-2x^2-2y^2}{4}=\frac{-(x-y)^2}{4}$$ then it is obvious that we have $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}\leq 0.$$ So  $$\frac{(x+y)^2}{4}\leq\frac{x^2+y^2}{2}.$$
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