Question

cho a,b,c>0 và a+b+c+ab+ac+bc=6abc

c/m\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)

Answer 1

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)

\(a+b+c+ab+ac+bc=6abc\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)

Hay \(x+y+z+xy+yz+xz=6\)

Cần chứng minh \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=x^2+y^2+z^2\ge3\)

Ta có : \(\left(x^2+1\right)+\left(y^2+1\right)+\left(z^2+1\right)\ge2\left(x+y+z\right)\) (BĐT Cosi)

\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\) (BĐT Cosi)

\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+xz\right)=12\)

\(\Rightarrow x^2+y^2+z^2\ge3\) (đpcm)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)