Question

Chứng minh: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>1\)với mọi a; b; c

Answer 1

\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

Answer 2

TA có :

   \(\frac{a}{b+c}>\frac{a}{a+b+c}\)   (1)

...........................  (2)

 \(\frac{c}{a+b}>\frac{c}{a+b+c}\left(3\right)\)

Từ (1) (2) và (3) =-> \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

=> ĐPCM