If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$. If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$. 
I know how to do the direct proof, but in this case, I want to try proving it by contradiction. I have tried manipulating the inequality $a + b < 2 \sqrt{ab}$ after making the assumption that $a,b >0$ to get a contadiction $a,b \leq 0$.
$\begin{align} a + b &< 2 \sqrt{ab} \\a^2-2ab+b^2 &< 0 \\(a-b)^2 &< 0\end{align}$
How do I show that $a,b \leq 0$?

\begin{align}
   a > 0, b > 0, a + b < 2\sqrt{ab}
   &\implies a - 2\sqrt{ab} + b < 0 \\
   &\implies (\sqrt a - \sqrt b)^2 < 0 \\
\end{align}
Which is a contradiction.
by the way
$a + b < 2 \sqrt{ab}\;$ implies  $a^2-2ab+b^2 > 0\;$, not what you said.
