Why is $xy ≤ 0.5(x^2+y^2)$?

Question

In order to prove a certain function in $\mathbb{R}^2$ is totally differentiable, I have to make the following step: $xy ≤ 0.5(x^2+y^2)$. However, I do not understand why this inequality holds. I do understand both $x$ and $y$ are smaller than the norm $\|x\|$, which would create the inequality $xy ≤ x^2+y^2$, but that expression is not small enough to justify the inequality.

Answer 1

Simply note that by SOS

$$xy ≤ 0.5(x^2+y^2)\iff x^2-2xy+y^2\ge0\iff(x-y)^2\ge0$$

Answer 2

A.M-G.M. Sneaks in everywhere!

Let $a = x^2; b= x^2$ then $\frac{x^2 + y^2}2 = \frac {a + b}2 \ge \sqrt{ab} = |xy| \ge xy$. (If $xy < 0$ then $0.5(x^2 + y^2) \ge 0 > xy$ is trivial.)

That's it.

If you are like me and never really trusted A.M-G.M. (I'm sure it was snuck in but outside agitators) theres always

$0.5(x^2 + y^2) \ge xy \iff$

$x^2 + y^2 \ge 2xy \iff$

$x^2 - 2xy + y^2 \ge 0 \iff$

$(x- y)^2 \ge 0 \iff$

the universe exists.

As the universe does exist, this is true.