Every concave function that is nonnegative on its domain is log-concave?

Question

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.

Answer 1

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

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

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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.

Answer 2

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