Question

CHO \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0,a+b+c=abc\)
CHỨNG MINH \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

Answer 1

I think that we have to prove \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)

We have \(a+b+c=abc\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

We have \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=0\)( Because \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\))

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=-2\)

So...