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 "="\(\Leftrightarrow a=b=c=\frac{1}{3}\))
Cho tớ sửa đề
tử của ba cái là mũ 2 lên hết nha
\(T=\frac{a^2}{a^2+8bc}+\frac{b^2}{b^2+8ca}+\frac{c^2}{c^2+8ab}\)
\(\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac+6\left(ab+bc+ac\right)}\)
\(\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+6.\frac{\left(a+b+c\right)}{3}^2}\)
\(=\frac{1}{1+\frac{6}{3}}=\frac{1}{3}\)
Dấu "=" xảy ra <=> a = b = c = 1/3
