Question

1. Cho a,b,c là độ dài 3 cạnh của tam giác. Cmr:

A= a / b+c-a + b /a+c-b + c/a+b-c >_ 3

Answer 1

Đặt \(b+c-a=x;a+c-b=y;a+b-c=z\)

thì \(2a=y+z;2b=x+z;2c=x+y\). Ta có:

\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)

\(=\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)

\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6\)

hay: \(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\left(đpcm\right)\)