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 $a,b \leq 0$?
Best Answer
You are dealing with real numbers... It is impossible that $(a-b)^2<0$ because squaring a quantity always results in another quantity that is greater than or equal to zero.
EDIT: You seem to believe that a very specific contradiction is needed to complete this proof. Namely, that $a,b \leq 0$ has to be reached before you can stop. That isn't how proof by contradiction works. As long as you get any form of contradiction, you can end the proof. The fact that you got $(a-b)^2<0$ is perfect, because it violates one of your assumptions. You assumed $a,b$ are real numbers in addition to assuming that $a,b >0$. The only way you can square $a-b$ and end up negative is if $a-b$ is a complex number. That violates the assumption that $a,b$ are real, hence you have a contradiction.