18
$\begingroup$

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 $xf(y)+yf(x)\leq 1$.

$\endgroup$
  • $\begingroup$ 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 '15 at 19:28
  • $\begingroup$ @JimmyK4542 You are right, I will edit the post. $\endgroup$ – Jolien May 22 '15 at 19:29
  • 4
    $\begingroup$ 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 '15 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 '15 at 19:44
22
$\begingroup$

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

$\endgroup$
  • $\begingroup$ This was IMC contest problem... $\endgroup$ – k1.M May 22 '15 at 19:54
  • 2
    $\begingroup$ It looks so easy this way! $\endgroup$ – Jolien May 22 '15 at 19:58
  • $\begingroup$ Definitely what the authors had in mind, but damn is that slick! $\endgroup$ – MichaelChirico May 22 '15 at 20:20
2
$\begingroup$

$$\int \limits _0 ^1 f(x) \Bbb d x = \frac 1 2 \int \limits _0 ^1 f(x) \Bbb d x + \frac 1 2 \int \limits _0 ^1 f(y) \Bbb d y = - \frac 1 2 \int \limits _{\frac \pi 2} ^0 f(\cos t) \sin t \Bbb d t + \frac 1 2 \int \limits _0 ^{\frac \pi 2} f(\sin t) \cos t \Bbb d t \le \frac 1 2 \cdot \frac \pi 2$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.