What is the proof that $(a+b)^2 >a^2 + b^2$?

Question

I would like to know if there is a theorem that proves that

$$(a+b)^2>a^2+ b^2$$

where $ab>0$

I am also wondering whether there is a name associated with this inequality.

Answer 1

By distributing out, you get that

$${(a+b)^2 = a^2 + 2ab + b^2 = (a^2 + b^2) + 2ab}$$

Now - since ${a,b>0}$, then ${2ab>0}$. In other words, ${(a+b)^2 = (a^2 + b^2) + \text{a little bit}}$. So

$${(a+b)^2 = (a^2 + b^2) + 2ab > a^2 + b^2}$$

Edit: Indeed as @KeithBackman has remarked - this argument still works even if both ${a,b<0}$ - the point is, we just need ${2ab>0}$ for this argument to work. If they are both negative, ${2ab}$ is still ${>0}$, so the argument holds still :)

Answer 2

We have

$$(a+b)^2=a^2+2ab+b^2>a^2+b^2$$

Answer 3

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

If $a$ and $b$ are of the same sign