Question

Cho x, y, z > 0 thỏa mãn: \(x^2+y^2+z^2=3\). Chứng minh: \(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\ge3\).

Answer 1

\(A=\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\)

\(\Leftrightarrow A^2=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow2A^2=\left(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}\right)+\left(\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\right)+\left(\frac{x^2y^2}{z^2}+\frac{z^2x^2}{y^2}\right)+12\)

\(\ge2\left(x^2+y^2+z^2\right)+12=6+12=18\)

\(\Rightarrow A\ge3\)