Question

cho a,b,c>0 thỏa mãn a+b+c=1.

tính P=\(\sqrt{\frac{\left(a+bc\right)\left(b+ca\right)}{c+ab}}+\sqrt{\frac{\left(c+ab\right)\left(b+ca\right)}{a+bc}}+\sqrt{\frac{\left(a+bc\right)\left(c+ab\right)}{b+ca}}\)

Answer 1

Ta có : \(\left\{{}\begin{matrix}a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\\b+ca=b\left(a+b+c\right)+ca=\left(b+c\right)\left(a+b\right)\\c+ab=c\left(a+b+c\right)+ab=\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)

Từ đó ta có :

\(P=\Sigma\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)\left(a+b\right)}{\left(a+c\right)\left(b+c\right)}}\)

\(P=\Sigma\sqrt{\left(a+b\right)^2}\)

\(P=\Sigma\left(a+b\right)\)

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

\(P=2\)