How do I prove $2xy ≤ x^2+y^2$?

Question

I'm quite new at calculus, how could I prove:

$$2xy ≤ x^2 + y^2$$

I just can end up in $(xy)^{1/2} ≤ (x+y)/2$. And even though I know that this is true too, I don't know how to prove it either. Thank you in advance :D

Answer 1

The trivial inequality $a^2 \geq 0$ is true for any $a$. Im particular, it must be true that

$$ (x-y)^2 \geq 0 $$

But, $(x-y)^2 = x^2 - 2xy + y^2 $. Therefore, the result follows.

Answer 2

$$ 0 \leq (x-y)^2 = x^2 + y^2 - 2xy $$ which yields the result.

Answer 3

$$x^2 \geq 0$$ $$(a \pm b)^2 \geq 0$$ $$a^2\pm 2ab+b^2 \geq 0$$ $$a^2+b^2 \geq \mp 2ab$$ In particular, using $(a-b)$, we get the following. $$\boxed{a^2+b^2 \geq 2ab}$$