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

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

• Can you prove that, in general, for $p$ and $q$ positive, $\frac{p}{q}+\frac{q}{p}\geq 2$? Now, can you use log rules to relate the given LHS to the fractions $\frac{p}{q}$? Mar 31, 2015 at 14:15
• Find the relation with $f(x)=x+\dfrac1x$ and study the latter.
– user65203
Mar 31, 2015 at 14:15

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.

• thanks for the compliment!
– cgo
Mar 31, 2015 at 14:42

By the base change formula for logs,

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

Thus

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

• I like the way to establish the inequation.
– user65203
Mar 31, 2015 at 14:18
• Downvotes for all answerers who stepped on the toes of the hints (by Michael and Yves), and did not wait for the OP to attempt the problem using the hints. Mar 31, 2015 at 14:19
• I actually didn't see the hints they gave until you mentioned it. No problem, I'll delete my answer.
– kobe
Mar 31, 2015 at 14:20
• No, don't do it Mar 31, 2015 at 14:20
• Ok then, I'll leave it up.
– kobe
Mar 31, 2015 at 14:23

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 $x \neq 1$:

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

• Downvoter care to comment? Mar 31, 2015 at 14:17
• You should also add that since, for this case, $x\neq \dfrac{1}{x}$, the equality case doesn't hold and we get a strict inequality. p.s- I didn't downvote your answer. :) Mar 31, 2015 at 14:18
• Beautiful proof Mar 31, 2015 at 14:20