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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](http://en.wikipedia.org/wiki/Subadditivity), see [here](http://math.stackexchange.com/a/80015/8297). The proof of the special case can be found [here](http://math.stackexchange.com/a/82627/8297). See also Exercise 16.6.4, [p.480](http://books.google.com/books?id=rqqvbKOC4c8C&pg=PA480), 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.2012-06-28

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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.

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    By homogeneity: if $ab=0$ it's clear, otherwise divide by $b^{1/p}$.2012-06-28