Moscow Middle School Math Olympiad Question | How to prove this inequality? | Mathematical Olympiad

How to prove this inequality which has two lograithmic functions? We use the properties of logarithmic functions, basic algebra and algebraic trick to complete the proof.

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[vcvartak]

Making (a+1/a) as perfect square is nice step, it gives instant result

[drozfarnyline]

you can use log base change property to get a=1/b immediately and then am gm to finish
Nice problem BTW
Thanks!

[josecarlosferreira]

1 / log2 (pi) = logpi (2)
1/logpi (2) = log2 (pi)
so,
1 / log2 (pi) + 1/logpi (2) = logpi (2) + log2 (pi)

By the AM-GM inequality, since the logs are positive:
(logpi (2) + log2 (pi) ) / 2 > sqrt(logpi (2) * log2 (pi)) *They cannot be equal because the terms are different.
(logpi (2) + log2 (pi) ) / 2 > 1 *1 / log2 (pi) = logpi (2)
(logpi (2) + log2 (pi) ) > 2

[jameswayne]

nice proof, you may also use the AM-GM for the second part of your proof.

[dandeleanu]

let Logπ2=y => y^2 + 1 > 2y => (y – 1)^2 > 0 is true because y <> 1

[j.r.]

This is a pretty weird proof (especially for middle school students!). What happens if you use [1/sqrt(a) + sqrt(a)]^2 in the last step, instead of the unexpected [1/sqrt(a) - sqrt(a)]^2? Of course, you would then need -2 instead of +2 to balance the equation. Does the proof still work as [1/sqrt(a) + sqrt(a)]^2 - 2 > 2?