Prove that $\int_0^1 \frac{dx}{f^2(x)+1} \le \frac{ \pi}{4}$ Let $f:[0,1] \to \mathbb{R}$ be a differentiable function, for which $f'(x) \ge 1 , \forall x\in [0,1]$, and $f(1)=1$. Prove that: $$\int_0^1 \frac{dx}{f^2(x)+1} \le \frac{ \pi}{4}$$
From the hyphotesis, we deduce that $f(0) \le 0$. This doesn't help very much, because if we write $$\int_0^1 \frac{dx}{f^2(x)+1} \le \int_0^1 \frac{f'(x)}{f^2(x)+1} dx =\arctan(f(x))|_0^1=\frac{ \pi }{4}-\arctan(f(0)) \ge \frac{ \pi }{4}$$
Another attempt is: We notice that $ \frac{ \pi }{4} = \int_0^1 \frac{dx}{x^2+1}$, so we only need to prove that $\int_0^1 \frac{dx}{f^2(x)+1} \le \int_0^1 \frac{dx}{x^2+1}$. This can be written as:
$$ 0 \le \int_0^1 \frac{(f(x)-x)(f(x)+x)}{(x^2+1)(f^2(x)+1)} dx $$
whichi, I think, isn't very easy to prove.
Also, I tried to use that the function $x \to f(x)-x$ is increasing (in fact, I used it when I proved that $f(0) \le 0$), but nothing.

As an addition to the answer above:
It is sufficient to look at linear function $f(x) = \frac{x-a}{1-a}$ with sufficient small $a \in [0, 1]$
$$ \int^1_0{\frac{dx}{1 + f^2(x)}} = \int^1_0{\frac{dx}{1 + (\frac{x-a}{1-a})^2}} = I $$ 
Then $ u = \frac{x-a}{1-a}$, $du = \frac{dx}{1-a}$
$$I = (1-a)\int^1_{\frac{-a}{1-a}}{\frac{du}{1 + u^2}} = \frac{\pi}{4} - a\frac{\pi}{4} + (1-a)\arctan(\frac{a}{1-a})$$
Now for sufficiently small $a$ (using first two terms in arctan series):$$I > \frac{\pi}{4} + a(1-\frac{\pi}{4} - \frac{a^2}{3(1-a)^2}) > \frac{\pi}{4}$$
Where "sufficiently small" means: $$a < \frac{\gamma}{1 + \gamma}, \gamma = \sqrt{3(1 - \frac{\pi}{4})}$$
