Maximum of $xy(1-\frac{z^2}{(x+y)^2})+yz(1-\frac{x^2}{(y+z)^2})+zx(1-\frac{y^2}{(z+x)^2})$

Question

Let $x,y,z>0$ and $x^2+y^2+z^2=1$. What is the maximum value of $$xy\left(1-\frac{z^2}{(x+y)^2}\right)+yz\left(1-\frac{x^2}{(y+z)^2}\right)+zx\left(1-\frac{y^2}{(z+x)^2}\right)?$$

We can get an upper bound of $1$ if we throw away the minus terms and use the fact that $xy+yz+zx\leq x^2+y^2+z^2$. However, when $x=y=z$ the value is $3/4$, while if some variable approaches zero the value is at most $1/2$.

Answer 1

For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3}{4}$.

We'll prove that it's a maximal value.

Indeed, we need to prove that $$\sum_{cyc}xy\left(1-\frac{z^2}{(x+y)^2}\right)\leq\frac{3}{4}(x^2+y^2+z^2)$$ or $$\sum_{cyc}\left(\frac{1}{4}z^2-\frac{z^2xy}{(x+y)^2}\right)\leq\sum_{cyc}(x^2-xy)$$ or $$\sum_{cyc}\frac{z^2(x-y)^2}{(x+y)^2}\leq2\sum_{cyc}(x-y)^2$$ or $$\sum_{cyc}(x-y)^2\left(2-\frac{z^2}{(x+y)^2}\right)\geq0.$$ Let $x\geq y\geq z$.

Hence, $$y^2\sum_{cyc}(x-y)^2\left(2-\frac{z^2}{(x+y)^2}\right)\geq y^2\sum_{cyc}(x-y)^2\left(1-\frac{z^2}{(x+y)^2}\right)=$$ $$=(x+y+z)y^2\sum_{cyc}\frac{(x-y)^2(x+y-z)}{(x+y)^2}\geq$$ $$\geq(x+y+z)\left(\frac{y^2(x-z)^2(x+z-y)}{(x+z)^2}+\frac{y^2(y-z)^2(y+z-x)}{(y+z)^2}\right)\geq$$ $$\geq(x+y+z)\left(\frac{x^2(y-z)^2(x-y)}{(x+z)^2}+\frac{y^2(y-z)^2(y-x)}{(y+z)^2}\right)=$$ $$=\frac{(x+y+z)(y-z)^2(x-y)(x^2(y+z)^2-y^2(x+z)^2)}{(x+z)^2(y+z)^2}=$$ $$=\frac{(x+y+z)z(y-z)^2(x-y)^2(2xy+xz+yz)}{(x+z)^2(y+z)^2}\geq0.$$ Done!