<s>[INST] Prove $\frac{1}{\log_2\pi }+\frac{1}{\log_5\pi}>2$ 
Knowing that $\pi^2 < 10$. Prove that:
  $$\frac{1}{\log_2\pi}+\frac{1}{\log_5\pi}>2.$$

I have tried to do this the following way:
$\log_2\pi+\log_5\pi>\frac{1}{2} \Leftrightarrow \log_2\pi+\frac{\log_2\pi}{\log_25}>\frac{1}{2}\Leftrightarrow 2\log_2\pi-\log_25>\frac{1}{2}$
Is this the right way to proceed?
 [/INST] Notice that:
\begin{align*}
\frac{1}{\log_2 \pi} + \frac{1}{\log_5 \pi}
&= \frac{1}{\left(\frac{\log_\pi \pi}{\log_\pi 2} \right)} + \frac{1}{\left(\frac{\log_\pi \pi}{\log_\pi 5} \right)} & \text{change of base} \\
&= \frac{\log_\pi 2}{\log_\pi \pi} + \frac{\log_\pi 5}{\log_\pi \pi} \\
&= \log_\pi 2 + \log_\pi 5 \\
&= \log_\pi 10 \\
&> \log_\pi (\pi^2) &\text{since $10 > \pi^2$ and $\log_\pi(x)$ is strictly increasing} \\
&= 2
\end{align*}
</s>