$xf(y)+yf(x)\leq 1$ for all $x,y\in[0,1]$ implies $\int_0^1 f(x) \,dx\leq\frac{\pi}{4}$ 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 $xf(y)+yf(x)\leq 1$.

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