Question

Answer 1

Áp dụng BĐT \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Ta có:

\(A^2=\left(\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}\right)^2\ge3\left(\frac{yz}{x}\cdot\frac{zx}{y}+\frac{yz}{x}\cdot\frac{xy}{z}+\frac{zx}{y}\cdot\frac{xy}{z}\right)=3\left(x^2+y^2+z^2\right)\)

\(\Rightarrow A^2\ge\frac94\Rightarrow A\ge\frac32\)

Dấu "=" xảy ra khi \(x=y=z=\frac12\)