WakeUp wrote: For all positive real numbers $x$ and $y$ let \[f(x,y)=\min\left( x,\frac{y}{x^2+y^2}\right) \] Show that there exist $x_0$ and $y_0$ such that $f(x, y)\le f(x_0, y_0)$ for all positive $x$ and $y$, and find $f(x_0,y_0)$. We'll show that $\boxed{f(x,y)\le f\left(\frac 1{\sqrt 2},\frac 1{\sqrt 2}\right)=\frac 1{\sqrt 2}}$ $\forall x,y>0$ If $x\le \frac 1{\sqrt 2}$, then $f(x,y)\le x\le \frac 1{\sqrt 2}$ If $x\ge\frac 1{\sqrt 2}$, then $y^2-\sqrt 2 y+x^2\ge 0$ (consider it as a quadratic in $y$) and so $\frac y{x^2+y^2}\le \frac 1{\sqrt 2}$ and so $f(x,y)\le\frac y{x^2+y^2}\le \frac 1{\sqrt 2}$ Q.E.D.

Hint: In polar, we have $f(r, \theta) = \text{min} \left( r \cos \theta, \; \frac{1}{r} \sin \theta \right)$