Functional Inequality from Vietnam

Setting $x=y=z=0$ one gets

$\frac{1}{2}f(0)+\frac{1}{2}f(0)-f^2(0)\ge\frac{1}{4}$

which is equivalent to $\displaystyle \left(f(0)-\frac{1}{2}\right)^2\le 0,$ so that $\displaystyle f(0)=\frac{1}{2}.$ Similarly, $\displaystyle f(1)=\frac{1}{2}.$

Now setting $y=z=0,$

$\displaystyle \frac{1}{2}f(0)+\frac{1}{2}f(0)-f(x)f(0)\ge\frac{1}{4}$

so that $\displaystyle f(x)\le\frac{1}{2}.$ Setting $y=z=1,$

$\displaystyle \frac{1}{2}f(x)+\frac{1}{2}f(x)-f(x)f(1)\ge\frac{1}{4}$

results in $\displaystyle f(x)\ge\frac{1}{2}.$ Thus $\displaystyle f(x)=\frac{1}{2},$ for all $x\in\mathbb{R}.$

This problem is from the Vietnamese National Olympiad in Mathematics for Secondary Schools, 1991. It was published in the Canadian Crux Mathematicorum (December 1992), with the solution in a later issue.