Bounding a simple ratio of bivariate functions 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 $\alpha \le \sqrt{\pi/2}$.

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 $LHS-RHS = (1-c)(x-y)^2 \ge 0$.
