$|xy-zw|\leq 1/4$ for $x+y+z+w=1$ Let $x+y+z+w=1$, $x,y,z,w\geq 0$. If $x=y=1/2,z=w=0$, we have $|xy-zw|=1/4$.
Is it true that $|xy-zw|\leq 1/4$ always?

Your assertion is indeed true !. By triangle inequality:
$|xy - zw| \leq |xy| + |zw| = xy + zw \leq \dfrac{(x+y)^2+(z+w)^2}{4} = \dfrac{f(u)}{4}$, whereas $f(u) = u^2+(1-u)^2, u = x+y$. Note that $0 \leq u \leq 1$, $f'(u) = 4u-2 = 0 \iff u = \dfrac{1}{2}$. Thus $f_{max} = \text{max}\{f(0),f(1/2),f(1)\} = 1$, and the claim is proved.
