Question

Cho các số dương x, y, z thoả mãn: \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\). Tìm giá trị lớn nhất của biểu thức: \(P=\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\)

Answer 1

Đặt \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\x+z=c\end{matrix}\right.\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

\(P=\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\)

\(\Rightarrow P\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\Rightarrow P\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)

\(\Rightarrow P_{max}=\frac{3}{2}\) khi \(a=b=c=\frac{1}{2}\Rightarrow x=y=z=\frac{1}{4}\)