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

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

Áp dụng BĐT cô-si cho hai số không âm ta có:

\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\sqrt{1}=2\)

\(\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\sqrt{1}=2\)

\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\sqrt{1}=2\)

Suy ra: \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+2+2+2=9\)

=>Điều phải chứng minh

đặt A= vế trái

nhân phá ngoặc A ta đc:

A=1+x/y+x/z+y/x+1+y/z+z/x+z/y+1

=3+(x/y+y/x)+(x/z+z/x)+(y/z+z/y)

áp dụng BĐT:a/b+b/a>=2

=>A>=3+2+2+2=9

vậy...

Dễ chứng minh được: \(xy\le\frac{x^2+y^2}{2};yz\le\frac{y^2+z^2}{2};zx\le\frac{z^2+x^2}{2}\)

Do đó \(xy+yz+zx\le x^2+y^2+z^2\Leftrightarrow3\left(xy+yz+zx\right)\le x^2+y^2+z^2+2xy+2yz+2zx\)

\(3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2\Leftrightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=3\)

\(\Rightarrow A_{max}=3\Leftrightarrow x=y=z=1\)

vãi cả 2015 ạ =))