Sửa đề: ab+bc+ca=1
Ta có: \(a^2+1=a^2+ab+bc+ca\)
\(=a\left(a+b\right)+c\left(b+a\right)\)
\(=\left(a+b\right)\left(a+c\right)\)
Ta có: \(b^2+1=b^2+ab+bc+ca\)
\(=b\left(b+c\right)+a\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
Ta có: \(c^2+1=c^2+ab+bc+ca\)
\(=c\left(c+a\right)+b\left(c+a\right)\)
\(=\left(a+c\right)\left(b+c\right)\)
Ta có: \(A=\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(=\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2\)
là bình phương của 1 số hữu tỉ(đpcm)
