Question

 

Answer 1

Dễ thấy : \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=xyz\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=x+y+z\)

Do đó : \(P=x+y+z+\dfrac{3}{x+y+z}\)

\(=\dfrac{x+y+z}{3}+\dfrac{3}{x+y+z}+\dfrac{2\left(x+y+z\right)}{3}\)

\(\ge2\sqrt{\dfrac{\left(x+y+z\right)}{3}\cdot\dfrac{3}{x+y+z}}+2\cdot\dfrac{3\sqrt[3]{xyz}}{3}\)

\(=2+2=4\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Vậy Min\(P=4\) khi \(x=y=z=\dfrac{1}{3}\)