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

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.

• 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. – Daniel Fischer Nov 20 '14 at 22:19

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.