Question

Cho tam giác ABC có độ dài ba cạnh là a, b, c. Chứng minh rằng :

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< 2\)

Answer 1

\(\left\{{}\begin{matrix}\frac{a}{b+c}< \frac{2a}{a+b+c}\\\frac{b}{c+a}< \frac{2b}{a+b+c}\\\frac{c}{a+b}< \frac{2c}{a+b+c}\end{matrix}\right.\) \(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \frac{2a+2b+2c}{a+b+c}=2\)