Problem 1052

. From the rearrangement inequality we have $a^{3}+b^{3} \geqslant a^{2} b+b^{2} a$ and thus

$$\frac{1}{a^{3}+b^{3}+a b c} \leqslant \frac{1}{a^{2} b+b^{2} a+a b c}=\frac{c}{a b c(a+b+c)}$$

Similarly, we have

$\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{c}{a b c(a+b+c)}+\frac{a}{a b c(a+b+c)}+\frac{b}{a b c(a+b+c)}$ which gives the right inequality. For the left inequality, we will use the inequalities of means. First, according to the inequality between harmonic mean and arithmetic mean, we have

$$\frac{3}{\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c}} \leqslant \frac{2 a^{3}+2 b^{3}+2 c^{3}+3 a b c}{3} .$$

Finally, according to the inequality between geometric mean and arithmetic mean, we have

$$3 a b c \leqslant a^{3}+b^{3}+c^{3}$$

which combined with the previous inequality gives

$$\frac{3}{a^{3}+b^{3}+c^{3}} \leqslant \frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}+\frac{1}{c^{3}+a^{3}+a b c}$$