Maximizing $ \frac {y + z + yz}{\left( 1 + y + z \right)^2} $ I was solving a WOOT Problem of the Day and I boiled the problem down to finding the maximum value of $$ \frac {y+z+yz}{\left( 1 + y + z \right)^2}. $$Intuitively, this seems to occur when $y=z$, since $y$ is symmetric, and after that, it is easy to find that $$ \frac {y+z+yz}{\left( 1 + y + z \right)^2} \le \frac {1}{3}. $$This seems "intuitive" but is not a proof. Any ideas for a formal proof? 
 A: Consider the new variables $u=y+z, v=yz$.  The old conditions of $y,z>0$ translate to $u,v>0$.  However, for any fixed $u$, the variable $v$ must lie in $(0,(u/2)^2]$, where the latter is achieved when $y=z$.
Thus, to maximize $\frac{u+v}{(1+u)^2}$, we set $v=(u^2/4)$, which gives the expression $$\frac{u+u^2/4}{(1+u)^2}$$
In particular this proves that $y=z$ at the maximum, which is indeed $1/3$.
A: Note that $$\frac{1}{3}-\frac{y+z+yz}{(1+y+z)^2}=\frac{(1/4) (2 y-z-1)^2+(3/4) (z-1)^2}{3(1+y+z)^2}\ge 0.$$
The equality is attained when $y=z=1$.
A: If you compute the derivatives of $$F=\frac {y+z+yz}{\left( 1 + y + z \right)^2}$$ you have, after simplifications $$F'_y=\frac{-y (z+1)+z^2+1}{(y+z+1)^3}$$ $$F'_z=-\frac{y^2-2 y (2 z+1)+(z-2) z+3}{(y+z+1)^4}$$ Using $F'_y=0$ gives $y=\frac{z^2+1}{z+1}$ and the numerator of $F'_z$ simplifies to $$(1+z+z^2)(z^2+2 z-1)=0$$ So the solutions are  $$y=2 \left(-1-\sqrt{2}\right),z=-1-\sqrt{2},F=\frac{1}{18} \left(1-3 \sqrt{2}\right)$$ $$y=2 \left(-1+\sqrt{2}\right),z=-1+\sqrt{2},F=\frac{1}{18} \left(1+3 \sqrt{2}\right)$$
