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.

  • $\begingroup$ f '(x) ≥ 1; f(x) = (1+a)x with a ≥ 0; x^2 + 1 ≤ f(x)^2 + 1. Then the integral of the inverse of x^2 + 1 which is arct x is ≥ the integral of the given inverse.Since the fisr is pi/4 END $\endgroup$
    – Piquito
    Apr 15 '15 at 17:21
  • $\begingroup$ @Luis: Why should we have f(x)=(1+a)x ??? $\endgroup$
    – npatrat
    Apr 16 '15 at 10:54
  • $\begingroup$ @vadim123 : Can you give me some details, because I don't know how to use your hint? Thank you! $\endgroup$
    – npatrat
    Apr 16 '15 at 10:57
  • $\begingroup$ @npatrat, oops you're right my hint was a stinker. $\endgroup$
    – vadim123
    Apr 16 '15 at 14:10

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: $$a < \frac{\gamma}{1 + \gamma}, \gamma = \sqrt{3(1 - \frac{\pi}{4})}$$


This inequality is not true. Here is a counterexample:

We choose $a\in(0,1)$ such that $\arctan a>a\frac{\pi}{4}$ ($a$ exists since we have $\displaystyle\lim_{h\to0}\frac{\arctan h}{h}=1$ and $\frac{\pi}{4}\in(0,1)$). Then we define the function $f$ on $[0,1]$ by $f(x)=\frac{x-a}{1-a}$, if $x\in[a,1]$ and $f(x)=x-a$, if $x\in[0,a]$.

$f$ satisfies the hypoteses and we have $$\int_0^1\frac{dx}{f^2(x)+1}=\int_0^a\frac{dx}{(x-a)^2+1}+\int_a^1\frac{dx}{(\frac{x-a}{1-a})^2+1}=\arctan a+(1-a)\frac{\pi}{4}>\frac{\pi}{4}$$

  • $\begingroup$ Your function is not differentiable at a!!! $\endgroup$
    – npatrat
    Apr 16 '15 at 10:51
  • $\begingroup$ The function $f$ is effectively not differentiable at the point $a$. In order to fix this problem, for $n\in\mathbb{N}$ such that $[a-1/n,a+1/n]\subset(0,1)$. we are going to replace $f$ on $[a-1/n,a+1/n]$ by a polynomial of degree $3$ so that the obtained function, denoted $f_n$, is differentiable on $[0,1]$. For $ n $ sufficiently large, one can check that $f_n$ satisfies the hypotheses and that $\int_0^1\frac{dx}{f_n^2(x)+1}>\pi/4$. $\endgroup$ Apr 16 '15 at 17:42

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