Question

Cho a, b là số dương . Chứng minh rằng

a) \(\frac{a}{b+c}\)+ \(\frac{b}{c+a}\)+\(\frac{c}{a+b}\)≥\(\frac{3}{2}\)

b)\(\frac{b+c}{a}\) + \(\frac{a+c}{b}\)+\(\frac{a+b}{c}\) ≥ 6

Answer 1

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\frac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)

\("="\Leftrightarrow a=b=c\)

\(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\left(\frac{b}{a}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2\sqrt{\frac{ab}{ab}}+2\sqrt{\frac{ac}{ac}}+2\sqrt{\frac{bc}{bc}}=2+2+2=6\)

\("="\Leftrightarrow a=b=c\)