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I have to show that the following inequation is true:

$\frac{\ln(x) + \ln(y)}{2} \leq \ln(\frac{x+y}{2})$

I transformed it into

$\frac{\ln(x \cdot y)}{2} \leq \ln(x+y) - \ln(2)$

because I thought that I better can show the inequation here, but I don't know how to proceed.

How can I proceed or am I completely wrong?

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    $\begingroup$ You should try to have something like $\ln(X) \leq \ln(Y)$, and then use the fact that $\ln$ is an increasing function. $\endgroup$ – Esperluet Dec 5 '15 at 20:46
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    $\begingroup$ Exponentiate both sides: $e^{\frac{\ln(x) + \ln(y)}{2}} \le e^{\ln(\frac{x+y}{2})}$ and see if you can proceed from here. $\endgroup$ – r9m Dec 5 '15 at 20:48
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How about we exponentiate both sides? We get

$$e^{(ln(x)+ln(y))/2}=e^{ln(x)/2}e^{ln(y)/2}=\sqrt{xy}$$

and

$$e^{ln((x+y)/2)}=(x+y)/2$$

Now the result is immediate per the AM-GM inequality.

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  • $\begingroup$ Wow! Brilliant idea. I would have never guessed. Thanks! $\endgroup$ – mgluesenkamp Dec 5 '15 at 20:59

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