Question

[Ťɧε⚡₣lαsɧ]

Cho a,b,c >0 thỏa mãn a+b+c=1 .Tính P=\(\sqrt{\frac{\left(a+bc\right)\left(b+ca\right)}{c+ab}}+\sqrt{\frac{\left(b+ca\right)\left(c+ab\right)}{a+bc}}+\sqrt{\frac{\left(c+ab\right)\left(a+bc\right)}{b+ac}}\)

Ai giúp với

Answer 1

[Nguyễn Việt Lâm]

\(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)

Tương tự: \(b+ca=\left(a+b\right)\left(b+c\right)\) ; \(c+ab=\left(a+c\right)\left(b+c\right)\)

\(\Rightarrow P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)