Problem 1806

75. (2007.04.13) Prove: (1) If $a \geqslant b \geqslant c$, then
$$\begin{array}{l} \left(a^{6} b^{4}+b^{6} c^{4}+c^{6} a^{4}\right)-\left(a^{4} b^{6}+b^{4} c^{6}+c^{4} a^{6}\right)= \\ \left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(a^{2}-c^{2}\right) \sum b^{2} c^{2} \geqslant 0 \end{array}$$

Thus, we have
$$\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}} \geqslant \frac{b^{4}}{a^{3}+b^{3}}+\frac{c^{4}}{b^{3}+c^{3}}+\frac{a^{4}}{c^{3}+a^{3}}$$

Therefore, we get
$$\begin{array}{l} 2\left(\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}}\right) \geqslant \\ \frac{a^{4}+b^{4}}{a^{3}+b^{3}}+\frac{b^{4}+c^{4}}{b^{3}+c^{3}}+\frac{c^{4}+a^{4}}{c^{3}+a^{3}} \geqslant \\ \frac{1}{2} \sum(a+b) \end{array}$$

Thus,
$$\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}} \geqslant \frac{a+b+c}{2}$$

The original proposition is proved.
(2) If $a \geqslant c \geqslant b$, since
$$\frac{a^{4}}{a^{3}+b^{3}}=a-\frac{b^{2}}{a+b}+\frac{b^{2}(a-b)^{2}}{a^{3}+b^{3}}$$

Therefore,
$$\begin{array}{l} \sum \frac{a^{4}}{a^{3}+b^{3}}-\frac{1}{2} \sum a=\sum \frac{b^{2}(a-b)^{2}}{a^{3}+b^{3}}-\left(\sum \frac{b^{2}}{a+b}-\frac{1}{2} \sum a\right)= \\ \sum \frac{b^{2}(a-b)^{2}}{a^{3}+b^{3}}-\frac{2 \sum b^{2}(b+c)(c+a)-\sum a \cdot \prod(b+c)}{2 \prod(b+c)}= \\ \sum \frac{b^{2}(a-b)^{2}}{a^{3}+b^{3}}-\frac{\sum b c\left[\left(\sum a\right)^{2}-3 \sum b c\right]+\left(\sum b c\right)^{2}-3 a b c \sum a}{2 \prod(b+c)}= \\ \sum \frac{b^{2}(a-b)^{2}}{a^{3}+b^{3}}-\frac{\sum b c \cdot \sum(b-c)^{2}+\sum a^{2}(b-c)^{2}}{4 \prod(b+c)}= \\ \sum \frac{b^{2}(a-b)^{2}}{a^{3}+b^{3}}-\sum \frac{(b-c)^{2}}{4(b+c)}= \\ \sum\left[\frac{c^{2}}{b^{3}+c^{3}}-\frac{1}{4(b+c)}\right](a-b)^{2}= \\ \sum \frac{\left(3 c^{2}+b c-b^{2}\right)(a-b)^{2}}{b^{3}+c^{3}} \end{array}$$

Since $a \geqslant c \geqslant b$, then $3 a^{2}+c a-c^{2} \geqslant 0$, thus, we only need to prove
$$\frac{\left(3 c^{2}+b c-b^{2}\right)(a-b)^{2}}{b^{3}+c^{3}}+\frac{\left(3 b^{2}+a b-a^{2}\right)(a-c)^{2}}{a^{3}+b^{3}} \geqslant 0$$

Also, since
$$\begin{array}{l} \left(a^{3}+b^{3}\right)^{2}\left(3 c^{2}+b c-b^{2}\right)+\left(a^{3}+c^{3}\right)\left(b^{3}+c^{3}\right)\left(3 b^{2}+a b-a^{2}\right) \geqslant \\ 3 c^{2}\left(a^{3}+b^{3}\right)^{2}-a^{2}\left(a^{3}+c^{3}\right)\left(b^{3}+c^{3}\right) \geqslant \\ 3 c^{2}\left(a^{3}+b^{3}\right)^{2}-a^{2}\left(a^{3}+c^{3}\right)\left(b^{3}+c^{3}\right)= \\ a^{5}\left(2 a c^{2}-b^{3}-c^{3}\right)+a^{2} c^{2}\left(a^{4}-c^{4}\right)+a^{2} b^{3} c^{2}(a-c)+5 a^{3} b^{3} c^{2}+3 b^{6} c^{2} \geqslant \\ 0 \end{array}$$

Therefore, we have
$$\frac{\left(a^{3}+b^{3}\right)\left(3 c^{2}+b c-b^{2}\right)}{b^{3}+c^{3}}+\frac{\left(a^{3}+c^{3}\right)\left(3 b^{2}+a b-a^{2}\right)}{a^{3}+b^{3}} \geqslant 0$$

Additionally, it is easy to prove
$$\frac{(a-b)^{2}}{a^{3}+b^{3}} \geqslant \frac{(a-c)^{2}}{a^{3}+c^{3}}$$

From equations (2) and (3), and noting that $3 c^{2}+b c-b^{2} \geqslant 0$, we get
$$\begin{array}{l} \frac{\left(3 c^{2}+b c-b^{2}\right)(a-b)^{2}}{b^{3}+c^{3}}+\frac{\left(3 b^{2}+a b-a^{2}\right)(a-c)^{2}}{a^{3}+b^{3}}= \\ \frac{\left(a^{3}+b^{3}\right)\left(3 c^{2}+b c-b^{2}\right)}{b^{3}+c^{3}} \cdot \frac{(a-b)^{2}}{a^{3}+b^{3}}+\frac{\left(a^{3}+c^{3}\right)\left(3 b^{2}+a b-a^{2}\right)}{a^{3}+b^{3}} \cdot \frac{(a-c)^{2}}{a^{3}+c^{3}} \geqslant \\ {\left[\frac{\left(a^{3}+b^{3}\right)\left(3 c^{2}+b c-b^{2}\right)}{b^{3}+c^{3}}+\frac{\left(a^{3}+c^{3}\right)\left(3 b^{2}+a b-a^{2}\right)}{a^{3}+b^{3}}\right] \cdot \frac{(a-c)^{2}}{a^{3}+c^{3}} \geqslant} \end{array}$$

Therefore, equation (1) holds.
In summary, the original proposition is proved.