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I want to show that if $f\colon [0,1]\to\mathbb{R}$ is continuous and $xf(y)+yf(x)\leq 1$ for all $x,y\in[0,1]$ then we have the following inequality: $$\int_0^1 f(x) \, dx\leq\frac{\pi}{4}.$$

The $\pi$ on the right hand side suggests we have to do something with a geometric function. Letting $f(x) = \frac{1}{1+x^2}$ we have equality but this function does not satisfy $xf(y)+yf(x)\leq 1$.

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  • $\begingroup$ Just a comment: the function $f(x) := \frac{1}{1+x^2}$ fails to satisfy $xf(y)+yf(x) \le 1$ for $x = 1$ and $y = 0.3$. $\endgroup$
    – JimmyK4542
    May 22, 2015 at 19:28
  • $\begingroup$ @JimmyK4542 You are right, I will edit the post. $\endgroup$
    – Jolien
    May 22, 2015 at 19:29
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    $\begingroup$ On the other hand, $f(x)=\sqrt{1-x^2}$ seems to do the job: the integral is $\frac{\pi}4$, since it is just the area of a quarter of a circle and the condition is satisfied, since $$x\sqrt{1-y^2}+y\sqrt{1-x^2}=\langle(x,y),(\sqrt{1-y^2},\sqrt{1-x^2})\rangle\le \sqrt{x^2+y^2}\sqrt{2-x^2-y^2}\leq\frac{(x^2+y^2)+(2-x^2-y^2)}2=1,$$ the first inequality being the Cauchy-Schwarz inequality and the second one being the inequality of geometric and aritmetic means. $\endgroup$
    – Dejan Govc
    May 22, 2015 at 19:40
  • $\begingroup$ @DejanGovc Yes, you are right. This also means the two proofs just given don't make sense. $\endgroup$
    – Jolien
    May 22, 2015 at 19:44

2 Answers 2

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Let $I = \displaystyle\int_{0}^{1}f(x)\,dx$.

Substituting $x = \sin \theta$ yields $I = \displaystyle\int_{0}^{\pi/2}f(\sin \theta)\cos\theta\,d\theta$.

Substituting $x = \cos \theta$ yields $I = \displaystyle\int_{0}^{\pi/2}f(\cos \theta)\sin\theta\,d\theta$.

Hence, $I = \dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\left[f(\sin \theta)\cos\theta+f(\cos\theta)\sin\theta\right]\,d\theta \le \dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}1\,d\theta = \dfrac{\pi}{4}$.

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  • $\begingroup$ This was IMC contest problem... $\endgroup$
    – k1.M
    May 22, 2015 at 19:54
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    $\begingroup$ It looks so easy this way! $\endgroup$
    – Jolien
    May 22, 2015 at 19:58
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    $\begingroup$ Definitely what the authors had in mind, but damn is that slick! $\endgroup$ May 22, 2015 at 20:20
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$$\int \limits _0 ^1 f(x) \Bbb d x = \frac 1 2 \int \limits _0 ^1 f(x) \Bbb d x + \frac 1 2 \int \limits _0 ^1 f(y) \Bbb d y = - \frac 1 2 \int \limits _{\frac \pi 2} ^0 f(\cos t) \sin t \Bbb d t + \frac 1 2 \int \limits _0 ^{\frac \pi 2} f(\sin t) \cos t \Bbb d t \le \frac 1 2 \cdot \frac \pi 2$$

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