Does this $(x+y)^2\ge 4xy$ hold for all real $x$ and $y$ - not only for non-negative?

Question

Does this $(x+y)^2\ge 4xy$ hold for all real $x$ and $y$ - not only for non-negative?

I'm pretty much sure it is:

Suppose above is not true, that is $(x+y)^2< 4xy$

Then we come to a contradiction $(x-y)^2< 0$, which means that our assumption is false => the opposite is true.

Answer 1

Your reasoning is right.

I think it's better the following.

Note $$(x+y)^2-4xy=(x-y)^2\geq0$$ What do you think?

Answer 2

Your method is correct, but I think it can be simplified. In fact, you can rewrite the inequality as: $$x^2+y^2+2xy-4xy>0\leftrightarrow (x-y)^2>0$$ which is always true $\forall x,y \in R$.

Answer 3

Because ${\left ( (x+ y)^{2}- 4xy \right )}'= 2(x- y)(1- {y}')$, so

$$(x+ y)^{2}- 4xy\geq (y+ y)^{2}- 4y^{2}= 0$$