Question

cho S=\(\frac{a}{b+c}\) +\(\frac{b}{c+a}\) +\(\frac{c}{a+b}\) biết a+b+c=7 và \(\frac{1}{a+b}\) +\(\frac{1}{b+c}\) +\(\frac{1}{c+a}\)=\(\frac{7}{10}\) .so sánh S và \(1\frac{8}{11}\)

Answer 1

Xét

 \(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=7\cdot\frac{7}{10}=\frac{49}{10}\)

\(\Leftrightarrow\frac{a+b}{a+b}+\frac{c}{a+b}+\frac{a+c}{a+c}+\frac{b}{a+c}+\frac{b+c}{b+c}+\frac{a}{b+c}=\frac{49}{10}\)

\(3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{49}{10}\Leftrightarrow S=\frac{19}{10}\)

Ta có:   \(1\frac{8}{11}=\frac{19}{11}\)

vì 19=19 ,\(\frac{1}{11}< \frac{1}{10}\)nên \(\frac{19}{11}< \frac{19}{10}\)

Vậy \(S>1\frac{8}{11}\)