Question

Cho x,y,z dương thỏa mãn : \(x+y+z=xyz\)

Tìm GTNN : \(Q=\frac{y+z}{x^2}+\frac{z+x}{y^2}+\frac{x+y}{z^2}\)

Answer 1

\(x+y+z=xyz\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)

\(Q=x\left(\frac{1}{y^2}+\frac{1}{z^2}\right)+y\left(\frac{1}{x^2}+\frac{1}{z^2}\right)+z\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)

\(Q\ge\frac{2x}{yz}+\frac{2y}{zx}+\frac{2z}{xy}=\frac{2\left(x^2+y^2+z^2\right)}{xyz}\ge\frac{2\left(xy+yz+zx\right)}{xyz}\)

\(Q\ge2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge2\sqrt{3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}=2\sqrt{3}\)

\(Q_{min}=2\sqrt{3}\) khi \(x=y=z=\sqrt{3}\)