Question

giúp mình bài 4 với;-;

Answer 1

Bài 5: 

-Đặt \(b+c-a=x;c+a-b=y;a+b-c=z\left(x,y,z>0\right)\).

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

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

\(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}=\dfrac{\dfrac{y+z}{2}}{x}+\dfrac{\dfrac{z+x}{2}}{y}+\dfrac{\dfrac{x+y}{2}}{z}=\dfrac{1}{2}.\left(\dfrac{y+z}{x}+\dfrac{z+x}{y}+\dfrac{x+y}{z}\right)=\dfrac{1}{2}.\left[\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\right]\ge\dfrac{1}{2}.\left(2+2+2\right)=3\)-Dấu bằng xảy ra \(\Leftrightarrow a=b=c\)