Question

[Đinh Tuấn Việt]

So sánh a/(b+c) + b/(a+c) + c/(a+b) với 1

Answer 1

[Minh Hiền]

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

Mà theo t/c dãy TSBN:

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

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