Question

[Hà Phương]

CMR: \(\left(x+y+z\right)^2\ge3\left(x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}\right)\) với \(x,y,z\ge0\)

Answer 1

[Mr Lazy]

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\)\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge xy+yz+zx+2\left(xy+yz+zx\right)=3\left(xy+yz+zx\right)\)

Áp dụng Côsi: 

\(xy+zx\ge2\sqrt{xy.zx}=2x\sqrt{yz}\)

Tương tự: \(xy+yz\ge2y\sqrt{zx};\text{ }yz+zx\ge2z\sqrt{xy}\)

\(\Rightarrow2\left(xy+yz+zx\right)\ge2\left(x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}\right)\)

\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\ge3\left(x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}\right)\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z\)