Ta có:
\(\frac{a^3b}{a^3+b^3}-\frac{ab^3}{a^3+b^3}=\frac{ab\left(a^2-b^2\right)}{a^3+b^3}=\frac{ab\left(a-b\right)}{a^2-ab+b^2}=\frac{a-b}{\frac{a}{b}+\frac{b}{a}-1}\ge\frac{a-b}{\frac{a}{b}+\frac{a}{a}-1}=\frac{b\left(a-b\right)}{a}\)
\(\frac{b^3c}{b^3+c^3}-\frac{bc^3}{b^3+c^3}=\frac{bc\left(b^2-c^2\right)}{b^3+c^3}=\frac{bc\left(b-c\right)}{b^2-bc+c^2}=\frac{b-c}{\frac{b}{c}+\frac{c}{b}-1}\ge\frac{b-c}{\frac{a}{c}+\frac{b}{b}-1}=\frac{c\left(b-c\right)}{a}\)
\(\frac{c^3a}{c^3+a^3}-\frac{ca^3}{c^3+a^3}=\frac{ca\left(c^2-a^2\right)}{c^3+a^3}=\frac{ca\left(c-a\right)}{c^2-ca+a^2}=\frac{c-a}{\frac{c}{a}+\frac{a}{c}-1}\ge\frac{c-a}{\frac{a}{c}+\frac{a}{a}-1}=\frac{c\left(c-a\right)}{a}\)
\(\Rightarrow\frac{a^3b}{a^3+b^3}-\frac{ab^3}{a^3+b^3}+\frac{b^3c}{b^3+c^3}-\frac{bc^3}{b^3+c^3}+\frac{c^3a}{c^3+a^3}-\frac{ca^3}{c^3+a^3}\ge\frac{b\left(a-b\right)+c\left(c-a\right)+c\left(b-c\right)}{a}=\frac{ab-b^2-ac+bc}{a}=\frac{\left(a-b\right)\left(b-c\right)}{a}\ge0\)
\(\Leftrightarrow\frac{a^3b}{a^3+b^3}+\frac{b^3c}{b^3+c^3}+\frac{c^3a}{c^3+a^3}\ge\frac{ab^3}{a^3+b^3}+\frac{bc^3}{b^3+c^3}+\frac{ca^3}{c^3+a^3}\left(đpcm\right)\)
