How can I prove that $2a^2 + 2b^2 > ab$

Question

I know that this statement is true, but I cannot figure out a way to actually prove it. $a$ and $b$ are both positive real numbers.

Answer 1

$2a^2 + 2b^2 > ab \iff 4a^2 + 4b^2 > 2ab \iff 3a^2 + 3b^2 + (a-b)^2 > 0$ for $a,b$ positive.

Answer 2

Dividing both sides by $ab$ gives $$2\Big( \frac{a}{b} + \frac{b}{a}\Big) > 1$$ Considering $\mu\equiv\frac{a}{b}$ and cleaning up a bit gives, $$\mu +\mu^{-1} >\frac{1}{2}$$ $a$ and $b$ were positive reals, so $\mu:(0, 1]\cup[1,+\inf)$. Thus, the sum of $\mu$ and it's inverse must be great than one. $\square$

Answer 3

Since $(a-b)^2>0$ for every real $a,b$ not both zero, we have that $$a^2+b^2>2ab>ab/2,$$ as wanted.