42. For any non-negative real numbers $x, y, z$, we have $x^{3}+y^{3}+z^{3}+c\left(x y^{2}+y z^{2}+z x^{2}\right) \geqslant(c+$ 1) $\left(x^{2} y+y^{2} z+z^{2} x\right)$. Find the maximum value of the real number $c$. (2008 Mongolian National Training Team Exam)
42 Assume $x=\min |x, y, z|, y=x+p, z=x+q(p, q \geqslant 0)$, substitute into the inequality

get
$$x^{3}+y^{3}+z^{3}+c\left(x y+y z+z x^{2}\right) \geqslant(c+1)\left(x^{2} y+y z+z^{2} x\right)$$
$$\begin{array}{l}
2\left(p^{2}-p q+q^{2}\right) x+p^{3}+q\left(q^{2}-p^{2}\right)+c p q(q-p) \geqslant \\
0(x, p, q \geqslant 0)
\end{array}$$

Notice that, for all $p, q \geqslant 0$, we have $p^{2}-p q+q^{2} \geqslant 0$. Therefore, for all $p ; q \geqslant 0$, it should be
$$p^{3} \mp q\left(q^{2}-p^{2}\right)+c p q(q-p) \geqslant 0$$
(1) When $q \geqslant p$, inequality (2) holds for all non-negative real numbers $c$.
(2) When $q1)$. Then the upper bound of $c$ is the minimum value of $f(u)$:
By differentiation, we know $\min _{u>1} f(u)=f\left(\frac{1+\sqrt{2}+\sqrt{2-\sqrt{2}-1}}{2}\right)=\frac{\sqrt{13+16 \sqrt{2}}-\Gamma}{2}=$
2. $4844 \cdots$.