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Let $\alpha,\beta,R>0$ and $p\in C^2([0,R),\Bbb R)$ satisfy $$p(x)\ge\alpha,\text{ and }(1+\beta)[p'(x)]^2\le p''(x)p(x),\quad\forall x\ge0$$ Hint: the function $p^{-\beta}$ is concave.[...]

I dont understand why $p^{-\beta}$ is a concave function. I can see that $p$ is convex because

$$(1+\beta)[p'(x)]^2\le p''(x)p(x)\implies (1+\beta)\frac{[p'(x)]^2}{p(x)}\le p''(x)\implies 0\le p''(x)$$

Maybe a silly question but I dont see where the assertion of the hint comes, can someone enlighten me please?

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hint: the function $f(x) = (p(x))^{-\beta} $ has $f''(x) = -\beta\cdot p^{-\beta-2}\left(p''p - (\beta+1)(p'(x))^2\right) \le 0$

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    Right! What a dumb... I suspected that the solution will be something simple after all. – 2017-01-09