Question

Cho a, b, c là 3 cạnh của một tam giác. CMR:

a, \(1< \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< 2\)

b, \(\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge3\)

Answer 1

b)\(\Sigma\frac{a}{b+c-a}=\Sigma\frac{a^2}{ab+bc-a^2}\)\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)(BĐT Svarxơ)\(\ge\frac{\left(a+b+c\right)^2}{\frac{2}{3}\left(a+b+c\right)^2-\frac{1}{3}\left(a+b+c\right)^2}\)(BĐT Bunhiacopxki)\(=3\)(đpcm)

a)\(\Sigma\frac{a}{b+c}=\Sigma\frac{a^2}{ab+bc}\)\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)\(\ge\frac{\left(a+b+c\right)^2}{\frac{2}{3}\left(a+b+c\right)^2}=1,5>1\)