\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)
=>\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
=>\(2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
=>\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)
=>\(\frac{c+a+b}{abc}=1\)
=> a+b+c=abc (đpcm)
Từ \(\left(1\right)\) suy ra : \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
Do \(\left(2\right)\) nên \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1,\) suy ra \(\frac{a+b+c}{abc}=1\\.\)
Do đó \(a+b+c=abc\)
