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let $F$ a closed convex subset of $\mathbb{R}^n$, let $x,y\in F$ and assume that for any $s\in[0,1]$ we have $f(s):=\mid sx+(1-s)y-z\mid\geq \mid y-z\mid$ why is it true that $\frac{\partial}{\partial s} f(s) |_{s=0}\geq 0$?

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Maybe because $f(s)\ge f(0)$ for every $s\ge0$? – Did Apr 18 '11 at 16:48
What is $z$ in your expression? – Willie Wong Apr 18 '11 at 16:57

$g(s):=\left(f(s)-|y-z|\right)\ge0$ for $s\ge0 $ and $g(0)=0$ so the minimum of $g$ happens at $s=0.$ Therefore, after $s=0$, $g$ increases; that's why the derivative must be non-negative.

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