Question

Với x,y,z dương thỏa mãn: \(x^2+y^2+z^2=\frac{2}{3}\left(x+y+z\right)+1\)

Tìm Gía trị nhỏ nhất:

\(P=\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\)

Answer 1

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

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

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

\(\Rightarrow\left(x+y+z+1\right)\left(x+y+z-3\right)\le0\)

\(\Rightarrow x+y+z-3\le0\)

\(\Rightarrow x+y+z\le3\)

\(P=\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\ge\frac{9}{xy+xz+yz}\ge\frac{9}{\frac{\left(x+y+z\right)^2}{3}}=\frac{27}{\left(x+y+z\right)^2}\ge\frac{27}{3^2}=3\)

\(\Rightarrow P_{min}=3\) khi \(x=y=z=1\)