Question

Cho a,b,c >0 thỏa mãn: ab+ bc+ca=1. Rút gọn biểu thức:

A= \(a\sqrt{\dfrac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}+b\sqrt{\dfrac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}+c\sqrt{\dfrac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}\)

Answer 1

Lời giải:

Do \(ab+bc+ac=1\) nên:

\(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)

\(b^2+1=b^2+ab+bc+ac=(b+a)(b+c)\)

\(c^2+1=c^2+ab+bc+ac=(c+a)(c+b)\)

Do đó:

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

\(=a\sqrt{\frac{(b+c)(b+a)(c+a)(c+b)}{(a+b)(a+c)}}+b\sqrt{\frac{(a+b)(a+c)(c+a)(c+b)}{(b+a)(b+c)}}+c\sqrt{\frac{(b+a)(b+c)(a+b)(a+c)}{(c+a)(c+b)}}\)

\(=a(b+c)+b(a+c)+c(a+b)=2(ab+bc+ac)=2\)

Vậy \(A=2\)