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We are given a positive, non-decreasing function $f$ defined on natural numbers with $f(0) = 0$. $f$ has a submodularity-like property:

$f(x+y) \leq f(x) + f(y) $

for all natural numbers $x$ and $y$.

Can we then show that for any natural numbers $x,y$, we have

$f(x) \geq \frac{x}{x+y}f(x+y)$?

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Well, I figured out that for fixed two number x,y the at least one of the inequalities must happens, i.e., for x or for y. –  checkmath Feb 20 '12 at 2:17
    
Functions and sequences with the above property are usually called subadditive. There is a chapter in the book An introduction to the theory of functional equations and inequalities by Marek Kuczma devoted to subadditive function on $\mathbb R$. I do not know off-hand about a reference that would contain a detailed study of subadditive sequences. –  Martin Sleziak Mar 18 '12 at 6:58

1 Answer 1

No.

For example pick $f(1) = f(2) = 1$ and $f(n)=2$ for $n \ge 2$. Then $1 = f(2) < \frac 23 f(3) = \frac 43$

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