Question

Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

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

Chứng minh a+b+c=abc

Answer 1

Lời giải:

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

\(\Rightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)

\(\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{4-2}{2}=1\) (do \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\) )

\(\Leftrightarrow \frac{a+b+c}{abc}=1\)

\(\Leftrightarrow a+b+c=abc\)

Do đó ta có đpcm.