Question

Cho a, b, c>0 thoả mãn a+b+c+ab+bc+ca=6abc

cmr:\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)

Answer 1

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

Ta có:

\(2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\frac{1}{a^2}+1+\frac{1}{b^2}+1+\frac{1}{c^2}+1\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Cộng vế với vế:

\(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+3\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=12\)

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