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.

$$
(a-b)^2 \geq 0 \\
(a-b)^2 = a^2 - 2ab + b^2 \\
a^2 + b^2 \geq 2ab \geq ab 
$$
