Prove that $\frac{1}{\log_{2}{\pi}}+\frac{1}{\log_{\pi}{2}}> 2$

Question

I have tried in many ways and i could not do it.

Answer 1

Remember that the sum of a positive number and its reciprocal is never less than 2. That is, $a + \dfrac{1}{a} \geq 2$. Equality happens when $a=1$.

In this case, notice that $log_{\pi} 2$ and $\log_2 \pi$ are reciprocals of each other. Hence, the conclusion.

Answer 2

For any $a, b > 0$, $\log_a b = {1 \over \log_b a}$. Hence the original inequality will be shown if we can demonstrate that for arbitrary $x > 0, x \neq 1$ that

$$x + {1 \over x} > 2 \ \ \ \ \ \ - (*) $$

We have $x \neq 1$ as $\log_2 \pi \neq 1$.

The relation (*) is true by the AM-GM inequality, strict inequality holding precisely because $x \neq 1$:

$${1 \over 2} \left( x + {1 \over x} \right) > \sqrt{x\cdot{1 \over x}} = 1$$

Answer 3

By the base change formula for logs,

$$\log_\pi 2 = \frac{1}{\log_2 \pi}.$$

Thus

$$\frac{1}{\log_2 \pi} + \frac{1}{\log_\pi 2} = \frac{1}{\log_2 \pi} + \log_2 \pi = \left(\sqrt{\frac{1}{\log_2 \pi}} - \sqrt{\log_2 \pi}\right)^2 + 2 > 2.$$