Question

a,b,c>0 TM a+b+c=1

cm\(\frac{c+ab}{a+b}+\frac{b+ac}{a+c}+\frac{a+bc}{b+c}=2\)

Answer 1

\(VT=\frac{c+ab}{a+b}+\frac{b+ac}{a+c}+\frac{a+bc}{b+c}\)

\(=\frac{c\left(a+b+c\right)+ab}{a+b}+\frac{b\left(a+b+c\right)+ac}{a+c}+\frac{a\left(a+b+c\right)+bc}{b+c}\)

\(=\frac{ac+bc+c^2+ab}{a+b}+\frac{ab+b^2+cb+ac}{a+c}+\frac{a^2+ab+ac+bc}{b+c}\)

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

Hình như là \(\ge2\) mới đúng bạn ạ :v

Answer 2

lm như thế nào nx ạk