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a KISS puzzle (Posted on 2017-02-27) Difficulty: 2 of 5
Prove:

logpie + logepi >2

Looks complicated?
It is not!

See The Solution Submitted by Ady TZIDON    
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Solution solution Comment 2 of 2 |
log{e}(pi) = 1/log{pi}(e)

So we need to show x + 1/x > 2 whenever x is not equal to 1. (Since pi and e are not equal, neither logarithm is 1.)

Take the function y = x + x^(-1). At 1 it has the value 2.

Its derivative is 1 - x^(-2), and at 1 that is zero, so the function is at a minimum or maximum.

The second derivative is 2x^(-3), which is 2 at x=1, which is positive, at any x, so x=1 is a minimum, and in fact the only relative extremum. Therefore any other value of x + 1/x is larger than 2 for any real x.

  Posted by Charlie on 2017-02-27 10:10:00
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