a) Assume the contrary. Then both $x,t$ are positive, or both are negative. If both are positive, then the sum is $> xt+yt+zt \ge 3xt \ge 1$, contradiction! Now, let $P(x,y,z,t)$ be the assertion that if $xy+xz+xt+yz+yt+zt = 1$, and $x\le y \le z \le t$, then $xt<\frac{1}{3}$. We proved it if $x$ and $t$ are positive. The negative case follows from $P(-t,-z,-y,-x)$. b) Let $x=y=z=r$, $t=\frac{1}{3r}-r$, with $0<r<\frac{1}{3}$. The inequality $x \le y \le z \le t$ can be easily seen to hold, and $xt=\frac{1}{3}-r^2$ approaches $\frac{1}{3}$ as $r$ goes to $0$. Therefore, $C=\frac{1}{3}$ is in fact the best constant. Cheers, Rofler