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...
 A: $\frac12 (x^2+y^2)-(\frac12 (x+y))^2$
$=\frac12 (x^2+y^2)-\frac14 (x^2+y^2+2xy)$
$=\frac14 (2x^2+2y^2-x^2-y^2-2xy)$
$=\frac14 (x^2+y^2-2xy)$
$=\frac14 (x-y)^2\ge 0$
A: This substitution is a nice way to do this:
$$x=r\sin\theta , y=r\cos\theta$$
So, our identity:
$$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
$$\frac{1}{4} (x^2+y^2+2xy) \leq \frac12(x^2 + y^2)$$
Applying our substitutions:
$$\frac{1}{4} (r^2+2\cdot r\sin\theta\cdot r\cos\theta) \leq \frac12(r^2)$$
$$\frac{r^2}{2}\cdot \frac{1+\sin2\theta}{2} \leq \frac{r^2}{2}$$
Now, if you note that the maximum value of $\sin 2\theta$ is $1$, then it becomes obvious.
Added: It could have also been done it this way:
$$\left(\frac{1}{2}(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
$$\frac{1}{2}\cdot\frac{1}{2}(r\cos\theta+r\sin\theta)^2 \leq \frac{r^2}{2}$$
$$\frac{r^2}{2}\cdot\frac{1}{2}(\cos\theta+\sin\theta)^2 \leq \frac{r^2}{2}$$
Now, note that $-\sqrt{2} \leq \cos\theta+\sin\theta = \sqrt{2}\cos(x+45^{\circ}) \leq \sqrt{2}$. Thus the maximum value of $(\cos\theta+\sin\theta)^2$ is $2$, and the result follow.
A: Proof by picture (angle H is 90°).

A: 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 $\frac 14$:
$$\frac 12 (x^2 + y^2) \ge \left(\frac 12 (x+y)\right)^2$$
A: $$(x-y)^2\geq0$$
$$2xy\leq x^2+y^2$$
$$x^2+y^2+2xy\leq 2x^2+2y^2$$
$$(x+y)^2\leq 2x^2+2y^2$$
A: 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  $$\frac{(x+y)^2}{4}\leq\frac{x^2+y^2}{2}.$$
A: Notice that 
$$(\frac{1}{2}(x+y))^2=\frac{1}{4}(x^2+2xy+y^2)\leq \frac{1}{4}(x^2+[x^2+y^2]+y^2)$$ (why?)
