Question

Cho 3 cạnh của 1 tam giác CMR

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

Answer 1

https://olm.vn/hoi-dap/detail/12121415915.html

vô đi rồi k cho mk

Answer 2

Ta co:

\(\frac{a^2}{ab+ca-a^2}+\frac{b^2}{ab+bc-b^2}+\frac{c^2}{ca+bc-c^2}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)}\)

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

Dau '=' xay ra khi \(a=b=c\)