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Let $f:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$, and $g:\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ be two functions. Suppose that $f(0) > g(0) = 0$, and $f$ is strictly increasing and concave, while $g$ is strictly increasing and strictly convex (so $f'(x)>0,\; f''(x) \leq 0,\; g'(x)>0$, and $g''(x) >0)$. Suppose that for some $x>0$ we have that $f(x)- g(x)>0$. Is it true that then for every $y \in [0,x]$ it must be that $f(y) - g(y) \geq 0$? Seems to be correct, at least for some special cases but, but I'm unable to prove it for the above outlined general case.

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$h(x) = f(x) - g(x)$ is strictly concave. If $h(0)\ge 0$ and $h(x) \ge 0$ for some $x > 0$ then $$ h(y) > \frac{x-y}{x} \, h(0) + \frac yx \, h(x) \ge 0 $$ for $y \in (0, x)$.

The monotony of $f$ and $g$ is not needed for this conclusion.

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  • $\begingroup$ Very nice. Thanks! $\endgroup$
    – Nidjsi
    Oct 17, 2018 at 7:18

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