Let $x,y,z$ be positive real numbers. Find the smallest positive $k$ such that $$\sum_{cyc}\left(\frac{(k-1)(z+x)}{y}-\frac{3}{2}\right)(x-y)^2\geq 0$$ holds true for $x,y,z$.
I proved it for $k=\frac{5}{2}$ as follows: WLOG, let $x\ge y\ge z$. Then $\frac{x+y}{z}-1\ge 2-1>0$ and $\frac{y+z}{x}-1+\frac{z+x}{y}-1>\frac{x}{y}+\frac{y}{x}-2\ge 0$. Since $(z-x)^2\ge (x-y)^2$, the result follows.
I don't know what's the smallest $k$.