Problem 1552

Solution. Notice that
$$\frac{2 a^{4}}{a^{3}+b^{3}}-a-\frac{3(a-b)}{2}=(a-b)\left(\frac{a\left(a^{2}+a b+b^{2}\right)}{a^{3}+b^{3}}-\frac{3}{2}\right)=\frac{2 b^{2}+a b-b^{2}}{3\left(a^{3}+b^{3}\right)}(a-b)^{2}$$

Therefore the inequality can be transformed to
$$S_{a}(b-c)^{2}+S_{b}(a-c)^{2}+S_{c}(a-b)^{2} \geq 0$$
in which the coefficients $S_{a}, S_{b}, S_{c}$ are
$$S_{a}=\frac{3 c^{2}+b c-b^{2}}{b^{3}+c^{3}} ; \quad S_{b}=\frac{3 a^{2}+c a-c^{2}}{c^{3}+a^{3}} ; \quad S_{c}=\frac{3 b^{2}+a b-a^{2}}{a^{3}+b^{3}}$$

The first case. If $a \geq b \geq c$, then clearly $S_{b} \geq 0$ and
$$\begin{aligned} S_{b}+2 S_{c} & =\frac{3 a^{2}+c a-c^{2}}{c^{3}+a^{3}}+\frac{2\left(3 b^{2}+a b-a^{2}\right)}{a^{3}+b^{3}} \geq \frac{3 a^{2}}{c^{3}+a^{3}}-\frac{2 a^{2}}{a^{3}+b^{3}} \geq 0 \\ a^{2} S_{b}+2 b^{2} S_{a} & =\frac{a^{2}\left(3 a^{2}+c a-c^{2}\right)}{c^{3}+a^{3}}+\frac{2 b^{2}\left(3 c^{2}+b c-b^{2}\right)}{b^{3}+c^{3}} \geq \frac{3 a^{4}}{c^{3}+a^{3}}-\frac{2 b^{4}}{c^{3}+b^{3}} \geq 0 \end{aligned}$$

So we conclude that
$$2 \sum_{c y c} S_{a}(b-c)^{2} \geq\left(S_{b}+2 S_{c}\right)(a-b)^{2}+(b-c)^{2}\left(2 S_{a}+\frac{a^{2}}{b^{2}} S_{b}\right) \geq 0$$

The second case. If $c \geq b \geq a$, then clearly $S_{a}, S_{c} \geq 0$ and
$$\begin{array}{l} S_{a}+2 S_{b}=\frac{3 c^{2}+b c-b^{2}}{b^{3}+c^{3}}+\frac{2\left(3 a^{2}+c a-c^{2}\right)}{c^{3}+a^{3}} \geq \frac{3 c^{2}+b c}{b^{3}+c^{3}}-\frac{2 c^{2}}{a^{3}+c^{3}} \geq 0 \\ S_{c}+2 S_{b}=\frac{3 b^{2}+a b-a^{2}}{a^{3}+b^{3}}+\frac{2\left(3 a^{2}+c a-c^{2}\right)}{c^{3}+a^{3}} \geq \frac{3 b^{2}}{a^{3}+b^{3}}-\frac{2 c^{2}}{c^{3}+a^{3}} \geq 0 \end{array}$$

We conclude that
$$2 \sum_{c y c} S_{a}(b-c)^{2} \geq\left(S_{a}+2 S_{b}\right)(b-c)^{2}+\left(2 S_{b}+S_{c}\right)(a-b)^{2} \geq 0$$

The proof is finished and the equality holds for $a=b=c$.