Maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$ What is the maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$? 

Wolfram Alpha plots this expression on a 3d graph, but I want to solve it algebraicly, by modifying the expression
My Attempts
1) add and substract 2 at the equation and we get $\frac{x+y+1}{y+1}+\frac{x+y+1}{x+1}$ and the numerator is same
=>failed
2) use AM-GM or Cauchy-Schwarz inequality
=>also failed

Since
$$
\delta\left(\frac{x}{y+1}+\frac{y}{x+1}\right)
=\left(\frac1{y+1}-\frac{y}{(x+1)^2}\right)\delta x
+\left(\frac1{x+1}-\frac{x}{(y+1)^2}\right)\delta y
$$
to get an interior critical point, we need
$$
\frac1{y+1}=\frac{y}{(x+1)^2}\quad\text{and}\quad\frac1{x+1}=\frac{x}{(y+1)^2}
$$
which have only the singular solution $x=y=-1$, which, even if allowed, are outside of $[0,1]\times[0,1]$. Therefore, we are searching for boundary critical points.
Due to the symmetry, we need only consider $x=0$ and $x=1$.
For $x=0$, we get $\frac{x}{y+1}+\frac{y}{x+1}=y$ and the maximum is at $y=1$.
For $x=1$, we get $\frac{x}{y+1}+\frac{y}{x+1}=\frac1{y+1}+\frac y2$. Then, the critical point is $y=\sqrt2-1$, which gives $\frac{x}{y+1}+\frac{y}{x+1}=\sqrt2-\frac12\lt1$. The endpoints $y\in\{0,1\}$ both give $\frac{x}{y+1}+\frac{y}{x+1}=1$.
Thus, the maximum is $1$.
