Áp dụng BĐT Svac - xơ
\(T=\frac{a}{a^2+8bc}+\frac{b}{b^2+8ca}+\frac{c}{c^2+8ab}\)
\(=\frac{a^2}{a^3+8abc}+\frac{b^2}{b^3+8abc}+\frac{c^2}{c^3+8abc}\ge\frac{\left(a+b+c\right)^2}{a^3+b^3+c^3+24abc}\)
Ta lại có : \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b+c\right)\left(ab+bc+ca\right)-3abc\)
\(\ge a^3+b^3+c^3+27\sqrt[3]{abc}.\sqrt[3]{\left(abc\right)^2}-3abc=a^3+b^3+c^3+24abc\)
Lúc đó : \(T\ge\frac{1}{a+b+c}=1\)
Dấu " = " xay ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)