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This is a statment from Wiki. I'm not sure why this is true:

If: $f(\theta x+(1-\theta y) \geq \theta f( x) + (1-\theta)f(y)$ And $f(\cdot) \geq 0$ then:

$$f(\theta x+(1-\theta y) \geq f( x)^{\theta}f(y)^{1-\theta}$$

I guess we should prove that:

$$\theta f( x) + (1-\theta)f(y) \geq f( x)^{\theta}f(y)^{1-\theta}$$

Don't have idea how to prove that.

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  • $\begingroup$ For that, forget about $f$ and prove $\theta a + (1-\theta)b \geqslant a^\theta b^{1-\theta}$ for $a,b \geqslant 0$. The case $\theta = \frac{1}{2}$ is a familiar inequality. $\endgroup$ – Daniel Fischer Nov 20 '14 at 22:19
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As $\log$ is a concave function, $$ \log (\theta a + (1-\theta)b )\ge \theta \log a + (1-\theta)\log b $$

Now with $$ a=f(x);b = f(y) $$it follows that

$$ \log f(\theta a + (1-\theta)b) \ge \theta \log f(x)+ (1-\theta)\log f(y) $$


More generally, if $f$ is concave with values in $D\subset \Bbb R$ and $g$ is concave increasing and define on $D$ then $g\circ f$ is concave as well: it is the same proof.

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This is the so-called "weighted arithmetic mean geometric mean inequality". See http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Weighted_AM.E2.80.93GM_inequality

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