Problem 1330

Analysis: Since the original inequality is not homogeneous, we first homogenize it, i.e., transform it into
$$\begin{array}{l} 0 \leqslant(x y+y z+z x)(x+y+z)-2 x y z \\ \leqslant \frac{7}{27}(x+y+z)^{3} . \\ \text { Let } f(x, y, z) \\ =(x y+y z+z x)(x+y+z)-2 x y z \\ =a g_{3,1}+b g_{3,2}+c g_{3,3} . \end{array}$$

To quickly calculate the undetermined coefficients, note that
$$\begin{array}{l} a=f(1,0,0), b=\frac{f(1,1,0)}{2} \\ c=f(1,1,1) \end{array}$$

Thus, we have $a=0, b=1, c=7$.
Proof: Note that
$$\begin{array}{l} (x y+y z+z x)-2 x y z \\ =(x y+y z+z x)(x+y+z)-2 x y z \\ = g_{3,2}+7 g_{3,3} \geqslant 0 . \\ \text { Then } \frac{7}{27}-(x y+y z+z x)+2 x y z \\ = \frac{7}{27}(x+y+z)^{3}-(x y+y z+z x) . \\ (x+y+z)+2 x y z \\ = \frac{7}{27} g_{3,1}+\frac{1}{27} g_{3,2} \geqslant 0 . \end{array}$$