, 2007

Solution:

Using the AM-GM inequality we have
$$\begin{aligned} & \sqrt{y z(x+m y)(x+n z)}=\sqrt{(x z+m y z)(x y+n y z)} \leq \frac{x y+x z+(m+n) y z}{2} \\ & \sqrt{x z(y+m x)(y+n z)}=\sqrt{(y z+m x z)(x y+n x z)} \leq \frac{x y+y z+(m+n) x z}{2} \\ & \sqrt{x y(z+m x)(z+n y)}=\sqrt{(y z+m x y)(x z+n x y)} \leq \frac{x z+y z+(m+n) x y}{2} \end{aligned}$$
Thus it is enough to prove that
$$\begin{aligned} x[x y+x z+(m+n) y z] & +y[x y+y z+(m+n) x z]+z[x y+y z+(m+n) x z] \\ \leq & \frac{3(m+n)}{4}(x+y)(y+z)(z+x), \end{aligned}$$
or
$$4[A+3(m+n) B] \leq 3(m+n)(A+2 B) \Leftrightarrow 6(m+n) B \leq[3(m+n)-4] A$$
where $A=x^{2} y+x^{2} z+x y^{2}+y^{2} z+x z^{2}+y z^{2}$, $B=x y z$.
Because $m+n \geq 2$ we obtain the inequality $m+n \leq 3(m+n)-4$. From AM-GM inequality it follows that $6 B \leq A$. From the last two inequalities we deduce that $6(m+n) B \leq[3(m+n)-4] A$. The inequality is proved.

Equality holds when $m=n=1$ and $x=y=z$.