Question

Chờ a,b,c là 3 cạnh của 1 tam giác.Cmr :

\(\dfrac{a}{b+c-a}\)+\(\dfrac{b}{a+c-b}\)+\(\dfrac{c}{a+b-c}\)>_3

Answer 1

\(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)

Ta gọi \(b+c-a=x;a+c-b=y;a+b-c=z\)

\(\Rightarrow2a=y+z;2b=x+z;2c=x+y\)

\(\Rightarrow2A=\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)

\(=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)\ge6\)\(\Rightarrow A\ge\dfrac{6}{2}=3\)