\(VT=\frac{a^3}{b^2+8}+\frac{b^3}{c^2+8}+\frac{c^3}{a^2+8}\)
\(=\frac{a^3}{b^2+ab+bc+ca}+\frac{b^3}{c^2+ab+bc+ca}+\frac{c^3}{a^2+ab+bc+ca}\)
\(=\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{c^3}{\left(c+a\right)\left(a+b\right)}\)
Áp dụng BĐT Cô si ta có :
\(\left\{{}\begin{matrix}\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{a+b}{8}+\frac{b+c}{8}\ge\frac{3a}{4}\\\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{b+c}{8}+\frac{c+a}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(c+a\right)\left(a+b\right)}+\frac{c+a}{8}+\frac{a+b}{8}\ge\frac{3c}{4}\end{matrix}\right.\)
\(\Rightarrow\frac{a^3}{\left(a+b\right)\left(b+c\right)}+\frac{b^3}{\left(b+c\right)\left(c+a\right)}+\frac{c^3}{\left(c+a\right)\left(a+b\right)}\ge\frac{a+b+c}{4}\ge\frac{\sqrt{3\left(ab+bc+ca\right)}}{4}=\frac{3}{4}\)
Vậy BĐT được chứng minh . Dấu = xảy ra khi \(a=b=c=1\)