# An inequality concerning concave functions

Let $I\subset\mathbb{R}^{\ge 0}$ be an interval and let $f:I\rightarrow\mathbb{R}^{\ge 0}$ be concave (and smooth enough). I'm wondering, weather the following inequality holds:

$$f(a+b) \le f(a) + f(b).$$

Is this true? I could not find a proof for it. I came to this question for the map $f(x) = x^{\frac{1}{p}}$, where $1\le p < \infty$. The inequality is true in this case, isn't it? I'm especially interested in this case.

I'd be thankful for any hint.

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Any concave function $f\colon\langle0,\infty)\to\mathbb R$ such that $f(0)=0$ is subadditive, see here. The proof of the special case can be found here. See also Exercise 16.6.4, p.480, in M. Kuczma: An introduction to the theory of functional equations and inequalities, where the function is defined on $(0,\infty)$ and the condition is: $\lim\limits_{x\to0^+}f(x)\ge0$ and $f$ is measurable. – Martin Sleziak Jun 28 '12 at 14:06
Awesome! Thank you! – Sh4pe Jun 28 '12 at 14:16

It is true with $f(x)=x^{1/p}$, $1\leq p<\infty$. To see that, note that we just have to deal with the case $b=1$. Then put $g(x):=x^{1/p}+1-(1+x)^{1/p}$ and show that this function is non-decreasing.
In the general case, if $0\in I$ and $f(0)=0$ it's true.
But the image of $f(x) = -1/x$ does not lie in the non-negative real numbers, as I stated above. Thank you very much in the $x^{1/p}$-case! :) – Sh4pe Jun 28 '12 at 14:10
Why is it sufficient to deal only with $b=1$? – Sh4pe Jun 28 '12 at 14:53
By homogeneity: if $ab=0$ it's clear, otherwise divide by $b^{1/p}$. – Davide Giraudo Jun 28 '12 at 14:54