Selection Examination A

The inequality is equivalent to $$\frac{x^2}{x+yz}+\frac{y^2}{y+zx}+\frac{z^2}{z+xy}\ge M.(1)$$ Since $x,y,z>0$, from Cauchy-Schwarz inequality we find $$\begin{array}{rl}& \left(\frac{x^2}{x+yz}+\frac{y^2}{y+zx}+\frac{z^2}{z+xy}\right)(x+yz+y+zx+z+xy)\ge (x+y+z{)}^2\\ \iff & \frac{x^2}{x+yz}+\frac{y^2}{y+zx}+\frac{z^2}{z+xy}\ge \frac{(x+y+z{)}^2}{x+y+z+xy+yz+zx}\\ \iff & \frac{x^2}{x+yz}+\frac{y^2}{y+zx}+\frac{z^2}{z+xy}\ge \frac{(x+y+z{)}^2}{x+y+z+1},\end{array}(2)$$ From $(x+y+z{)}^2\ge 3(xy+yz+zx)$, it follows that: $x+y+z\ge \sqrt{3}$, where the equality holds when $x=y=z=\sqrt{3}/3$. Now we consider the function $f(u)=\frac{u^2}{u+1}$, $u\ge \sqrt{3}$, which is strictly increasing, because for $u>v\ge \sqrt{3}$ we have $f(u)>f(v)$. In fact we have $$\begin{array}{rl}f(u)>f(v)& \iff \frac{u^2}{u+1}>\frac{v^2}{v+1}\\ & \iff u^{2}v+u^2-u{v}^2-v^2>0\\ & \iff (u-v)(uv+u+v)>0\\ & \iff u-v>0,\text{ since }uv+u+v>0.\end{array}$$ Therefore $f(u)\ge f(\sqrt{3})=\frac{3}{\sqrt{3}+1}$, and hence from inequality (2) we have: $$\frac{x^2}{x+yz}+\frac{y^2}{y+zx}+\frac{z^2}{z+xy}\ge \frac{(x+y+z{)}^2}{x+y+z+1}\ge \frac{3}{\sqrt{3}+1},$$ for all $x,y,z>0$ with $xy+yz+zx=1$. Hence: $M=\frac{3}{\sqrt{3}+1}$.