Question

Cho các số dương x,y,z thỏa mãn: x(x+1) + y(y+1) + z(z+1) \(\le18\)

Tìm GTNN: \(P=\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\)

Answer 1

\(18\ge x^2+y^2+z^2+x+y+z\ge\frac{1}{3}\left(x+y+z\right)^2+x+y+z\)

\(\Leftrightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-54\le0\)

\(\Leftrightarrow\left(x+y+z+9\right)\left(x+y+z-6\right)\le0\)

\(\Leftrightarrow x+y+z-6\le0\)

\(\Leftrightarrow x+y+z\le6\)

Do đó:

\(P\ge\frac{9}{2\left(x+y+z\right)+3}\ge\frac{9}{2.6+3}=\frac{3}{5}\)

Dấu "=" xảy ra khi \(x=y=z=2\)