Question

Cho 3 số dương a,b,c thoả mãn a+b+c=6 . Tìm GTLN của biểu thức:

\(P=\frac{ab}{6-c}+\frac{bc}{6-a}+\frac{ac}{6-b}\)

Answer 1

\(P=\frac{ab}{6-c}+\frac{bc}{6-a}+\frac{ac}{6-b}\)

\(P=\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\)

Ta có: \(\hept{\begin{cases}ab\le\frac{\left(a+b\right)^2}{4}\\bc\le\frac{\left(b+c\right)^2}{4}\\ac\le\frac{\left(a+c\right)^2}{4}\end{cases}}\)(bđt AM-GM)

\(\Rightarrow P\le\frac{\left(a+b\right)^2}{4\left(a+b\right)}+\frac{\left(b+c\right)^2}{4\left(b+c\right)}+\frac{\left(a+c\right)^2}{4\left(a+c\right)}=\frac{a+b+b+c+a+c}{4}=3\)

\("="\Leftrightarrow a=b=c=2\)