Prove that if {a;b} $\in \mathbb R^+$ then $a^2+b^2>ab$ I have tried factoring it already, but it doesn't seem to evolve much:
First I multiply each side by $2$:
$ 2(a^2+b^2)>2ab$
Then I substitute using the relation $(a+b)^2=a^2+2ab+b^2$ and it becomes:
$2(a^2+b^2)>(a+b)^2 - (a^2+b^2)$
and then:
$3(a^2+b^2)>(a+b)^2$
And that's pretty much it, I'm stuck.

We have that
$$0\le (a-b)^2=a^2+b^2-2ab.$$ Thus we have
$$2ab\le a^2+b^2.$$
Now, if $ab$ is positive then
$$ab< 2ab\le a^2+b^2.$$ And if $ab$ is negative then
$$ab< 0 <a^2+b^2.$$
