\(a,b,c>0and\left(a+b\right)\left(b+c\right)\left(a+c\right)=1\).Tìm max của \(ab+bc+ac\)
We have \(\left(a+b\right)\left(b+c\right)\left(a+c\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+ab+ac\right)\)
\(\Leftrightarrow1\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ac\right).\)
\(\Leftrightarrow\frac{9}{8}\ge\left(a+b+c\right)\left(ab+bc+ac\right)\ge\sqrt{3\left(ab+bc+ac\right)^3}.\)
\(\Leftrightarrow\frac{81}{64}\ge3\left(ab+bc+ac\right)^3\)
\(\Leftrightarrow\frac{27}{64}\ge\left(ab+bc+ac\right)^3\)
\(\Leftrightarrow\frac{3}{4}\ge ab+bc+ac\)
Vậy Max là \(\frac{3}{4}.\)Dấu bằng xảy ra khi \(a=b=c=\frac{1}{2}.\)
