\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\)
\(\Leftrightarrow\frac{ab+ac+bc}{abc}=\frac{1}{abc}\)
\(\Rightarrow ab+ac+bc=1\)
Ta có :
\(1+a^2=ab+ac+bc+a^2=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+b\right)\)
\(1+b^2=ab+ac+bc+b^2=a\left(b+c\right)+b\left(b+c\right)=\left(a+b\right)\left(b+c\right)\)
\(1+c^2=ab+ac+bc+c^2=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) là bình phương của 1 số hữu tỉ (ĐPCM)