Derivative with integral inequality proof

Question

let $f(x)$ be second derivative on $[0,1]$,and $$f''(x)\ge 0,f'(x)<0,\forall x\in[0,1], f(0)=0,f(1)=-1$$

show that $$\int_{0}^{1}\sqrt{\dfrac{f'^2(x)+1}{|f(x)|}}dx\le 2\sqrt{2}$$

Answer 1

This question is not well-posed.

Let $f(x)=x^{10}-2x$,satisfying $$f''(x) \ge 0, \forall x \in [0,1],f(0)=0,f(1)=-1$$but$$\int_{0}^{1}\sqrt{\dfrac{f'^2(x)+1}{|f(x)|}}dx=\int_{0}^{1}\sqrt{\dfrac{(10x^9-2)^2+1}{|x^{10}-2|}}dx=3.28097>2\sqrt2$$

Actually the supremum of the functional is $\infty$.