amparvardi wrote: Let $x \geq y \geq z$ be real numbers such that $xy + yz + zx = 1$. Prove that $xz < \frac 12.$ Is it possible to improve the value of constant $\frac 12 \ ?$ let $y=z=t,x=\frac{1-t^2}{2t},t\ne0$, satisfy $x+y+z=1$. then $xz=\frac{1}{2}-\frac{t^2}{2}$. let $t\to+0$ satisfy $x\ge y \ge z$ and $xz\to\frac 12$, so the constant $\frac 12$ is the best.

As for the inequality itself, $0\leq (x-y)(y-z) = xy + yz - y^2 - xz = 1 - y^2 - 2xz$, whence $xz \leq \dfrac {1-y^2} {2} < \dfrac {1} {2}$, since $y=0$ would imply $x\geq 0\geq z$, so $xz \leq 0$, and also $xz=1$, contradiction.