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I wonder what is the functions family that satisfies the following inequality:

$$\int_0^1 \frac{dx}{1+f^{2}(x)} <\frac{f(1)}{f'{(1)}}$$

This inequality seems to be a very interesting inequality, but not sure when it works and when not. For example, it works if i take $f(x)=e^x$ that is $\int_0^1 \frac{dx}{1+e^{2x}} = 0.28 < 1.$

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Why is it an interesting inequality? –  Mercy Jul 9 '12 at 11:49
    
@Mercy: because there are many calculus problems where you need to prove that a similar integral to this one is smaller than a certain value. It would help me solve many of these problems. Usually, it's rather hard to prove such inequalities and such a generalization is welcome ( of course, for some functions only). –  Chris's sis Jul 9 '12 at 11:51
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Here's my first though: If we assume that: $f'$ is continuous and not increasing; $f'(1) \neq 0$; $f(0) = 0$; $f(1) \ge 0$, than: $$\int_0^1 \frac{f'(x)}{1 + f^2 (x)} \cdot \frac{dx}{f'(x)} \le \frac{\arctan f(1)}{f'(1)} \le \frac{f(1)}{f'(1)}$$ –  qoqosz Jul 9 '12 at 11:54
    
You could try to find functions $u$ satisfying $$J(u):=\int_0^1[1+u^2(x)]^{-1}dx-u(1)/u'(1)=a$$ where $a \in \mathbb{R}$. –  Mercy Jul 9 '12 at 11:56
    
@qoqosz: it seems your way works just fine. –  Chris's sis Jul 9 '12 at 12:01

1 Answer 1

up vote 2 down vote accepted

If we assume that: $f'$ is continuous and not increasing; $f'(1) \neq 0$; $f(0) = 0$; $f(1) \ge 0$, than: $$\int_0^1 \frac{f'(x)}{1 + f^2 (x)} \cdot \frac{dx}{f'(x)} \le \frac{\arctan f(1)}{f'(1)} \le \frac{f(1)}{f'(1)}$$

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thank you! (+1) and answer accepted –  Chris's sis Sep 16 '12 at 9:43

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