# Simple Concave Function Question

Suppose we have a function $$f:[0,\infty)\longrightarrow [0,\infty)$$ that is concave and $$a > b>0$$. Then given a constant $$c>0$$ I claim that $$f(a+c) - f(a) \le f(b+c) - f(b)$$.

If I draw a picture, this statement seems obvious, but I can't seem to find a simple proof to actually show that this is the case!

Any help would be greatly appreciated. Thanks.

• i think that the place to start is en.wikipedia.org/wiki/Concave_function Commented Aug 4, 2020 at 1:53
• I also think that it is more appropriate (in this case) for the OP to show work before being provided an answer. To the OP: please spend 30 minutes to an hour trying (on scratch paper) to solve the problem. Then, edit your posting to show your work. Commented Aug 4, 2020 at 2:06

Hint: On one hand your inequality is same as $$\frac{f(a+c)-f(a)}{c} \leqslant \frac{f(b+c)-f(b)}{c}$$ and on second hand, if concave function have second derivative, then it is non-positive i.e. first derivative is not increasing.
Addition. In more general case, without derivatives, definition of concavity we can rewrite for any $$x_1 as $$\frac{f(x)-f(x_1)}{x-x_1} \geqslant \frac{f(x)-f(x_2)}{x_2-x}$$.
• the statement is also true when $f$ is not differentiable Commented Aug 4, 2020 at 14:57