Bounding a simple ratio of bivariate functions

Question

For any $x,y \ge 0$, define $f(x,y) := x^2+y^2 + 2c xy$, where $c := \sqrt{2/\pi}$. It is clear that $$ \sqrt{2/\pi}\cdot (x+y)^2 \le f(x,y) \le (x+y)^2. \tag{1} $$

This is an immediate consequence of the fact that $c \le 1 \le 1/c$.

Question. What is a good upper-bound for $\alpha := \sup_{x,y}\dfrac{(x+y)^2}{f(x,y)}$ ?

Note that (1) gives the trivial upper-bound $\alpha \le \sqrt{\pi/2}$.

Answer 1

Since $c<1$ and $f(x,y) =(x+y)^2-2xy(1-c)$ we have $$\frac1\alpha := \inf_{x,y}\dfrac{f(x,y)}{(x+y)^2}= 1-(1-c)\sup_{x,y}\frac{2xy}{(x+y)^2}$$

It remains to compute $$\sup_{x,y}\frac{2xy}{(x+y)^2}= \sup_{x,y}\frac12\frac{(x+y)^2-(x-y)^2}{(x+y)^2}= \frac12-\frac12\inf_{x,y}\frac{(x-y)^2}{(x+y)^2}=\frac12$$

Indeed since $x,y\geq 0$, taking $x=y=1$ we find that $$\inf_{x,y}\frac{(x-y)^2}{(x+y)^2}=\min_{x,y}\frac{(x-y)^2}{(x+y)^2}=0$$ Thus it follows that $$\frac1\alpha := \inf_{x,y}\dfrac{f(x,y)}{(x+y)^2}= 1-(1-c)\sup_{x,y}\frac{2xy}{(x+y)^2}= \frac{1+c}{2}$$ that is taking $x=y=1$ we find $$\alpha=\sup_{x,y}\dfrac{(x+y)^2}{f(x,y)}=\max_{x,y}\dfrac{(x+y)^2}{f(x,y)}= \frac{2}{1+c}\leq \frac1c.$$

Answer 2

It is natural to guess that the maximum is achieved when $x=y$, that is: \begin{equation} \frac{(x+y)^2}{x^2+y^2+2cxy}\le \frac{2}{1+c}, \end{equation} which is equivalent to \begin{equation} 2(x^2+y^2+2cxy) \ge (1+c)(x+y)^2, \end{equation} which is true because $LHS-RHS = (1-c)(x-y)^2 \ge 0$.