\(a+b+c=\frac{1}{abc}\)
\(\Leftrightarrow abc\left(a+b+c\right)=1\)(*)
\(\Leftrightarrow a^2bc+ab^2c+abc^2=1\)
Ta có :
\(1+b^2c^2=a^2bc+ab^2c+abc^2+b^2c^2\)
\(=abc\left(a+b\right)+bc^2\left(a+b\right)\)
\(=bc\left(a+b\right)\left(a+c\right)\)
Tương tự ta cũng có \(1+a^2c^2=ac\left(a+b\right)\left(b+c\right)\)
Khi đó : \(\left(1+b^2c^2\right)\left(1+a^2c^2\right)=abc^2\left(a+b\right)^2\left(b+c\right)\left(a+c\right)\)(1)
Xét \(c^2+a^2b^2c^2\)
\(=a^2b^2c^2+\frac{abc^3}{abc}\)
\(=a^2b^2c^2+abc^3\left(a+b+c\right)\)( theo giả thiết )
\(=a^2b^2c^2+a^2bc^3+ab^2c^3+abc^4\)
\(=abc^2\left(ab+bc+ca+c^2\right)\)
\(=abc^2\left(b+c\right)\left(a+c\right)\)(2)
Từ (1) và (2) ta suy ra :
\(\sqrt{\frac{\left(1+b^2c^2\right)\left(1+a^2c^2\right)}{c^2+a^2b^2c^2}}=\sqrt{\frac{abc^2\left(a+b\right)^2\left(b+c\right)\left(a+c\right)}{abc^2\left(b+c\right)\left(a+c\right)}}\)
\(=\sqrt{\left(a+b\right)^2}=\left|a+b\right|=a+b\)( vì \(a,b\in Z^+\) )
Ta có đpcm.
