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

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    $\begingroup$ 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 $\frac{p}{q}$? $\endgroup$ – Michael Burr Mar 31 '15 at 14:15
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    $\begingroup$ Find the relation with $f(x)=x+\dfrac1x$ and study the latter. $\endgroup$ – Yves Daoust Mar 31 '15 at 14:15
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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.

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  • $\begingroup$ thanks for the compliment! $\endgroup$ – cgo Mar 31 '15 at 14:42
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By the base change formula for logs,

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

Thus

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

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  • $\begingroup$ I like the way to establish the inequation. $\endgroup$ – Yves Daoust Mar 31 '15 at 14:18
  • $\begingroup$ 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. $\endgroup$ – GEdgar Mar 31 '15 at 14:19
  • $\begingroup$ I actually didn't see the hints they gave until you mentioned it. No problem, I'll delete my answer. $\endgroup$ – kobe Mar 31 '15 at 14:20
  • $\begingroup$ No, don't do it $\endgroup$ – Voyager Mar 31 '15 at 14:20
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    $\begingroup$ Ok then, I'll leave it up. $\endgroup$ – kobe Mar 31 '15 at 14:23
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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 $x \neq 1$:

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

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  • $\begingroup$ Downvoter care to comment? $\endgroup$ – Simon S Mar 31 '15 at 14:17
  • $\begingroup$ 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. :) $\endgroup$ – Prasun Biswas Mar 31 '15 at 14:18
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    $\begingroup$ Beautiful proof $\endgroup$ – Voyager Mar 31 '15 at 14:20

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