Prove $\frac{1}{\log_2\pi }+\frac{1}{\log_5\pi}>2$

Question

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?

Answer 1

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*}

Answer 2

$$\frac{1}{\log_2\pi}+\frac{1}{\log_5\pi}=\log_{\pi}2+\log_{\pi}5=\log_{\pi}10>\log_{\pi}\pi^2=2$$

Answer 3

Knowing only natural logarithms and using $$\log_a(b)=\frac{\log (b)}{\log (a)}$$ we then have $$\frac{1}{\log_2\pi}+\frac{1}{\log_5\pi}=\frac{\log (2)}{\log (\pi )}+\frac{\log (5)}{\log (\pi )}=\frac{\log (10)}{\log (\pi )}=\frac{2\log (10)}{\log (\pi^2 )} >\frac{2\log (10)}{\log (10 )}=2$$