Prove an integral inequation. Please help!! [duplicate]

Let $f$ is two differentiable function on $[0,1]$ such that $f(0)=1$, $f(1)=0$, $f'(x)\le 0$, $f''(x)\ge0$

Show that

$$\int_{0}^{1}x \sqrt{1+\{f'(x)\}^2}dx\le\ \frac{1}{\sqrt{2}}$$

How do have it? Thank you.

By integration by parts the given integral equals

$$\left[\frac{x^2}{2}\sqrt{1+f'(x)^2}\right]_{0}^{1}-\int_{0}^{1}\frac{x^2}{2}\cdot\frac{-2f'(x)f''(x)}{2\sqrt{1+f'(x)^2}}\,dx $$ or $$ \frac{1}{2}\sqrt{1+f'(1)^2}+\frac{1}{2}\int_{0}^{1}\frac{x^2 f'(x) f''(x)}{\sqrt{1+f'(x)^2}}\,dx\leq \frac{1}{2}\sqrt{1+f'(1)^2} $$ since $f'(x)\leq 0$ and $f''(x)\geq 0$.
$|f'(1)|\leq 1$ holds by convexity, so we are done.