I have to prove by contraposition that if $a \neq b$, then $(a + b)/2 > \sqrt{ab}$, where $a$ and $b$ are both positive real numbers.

So I need to prove that if $(a + b)/2 \leq \sqrt{ab}$, then $a=b$.

\begin{align}
\left(\frac{a+b}{2}\right)^2 &\leq ab \\
(a + b)^2 &\leq 4ab
\end{align}

I don't know if I started correctly, or what I should do next to get $a=b$.