Đề bài cậu ghi thiếu điều kiện phải không ? Điều kiện : a,b,c >0
Giải
\(\left|\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-\left(\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a}\right)\right|< 1\)
\(\Rightarrow-1< \left|\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{a}{c}-\dfrac{c}{b}-\dfrac{b}{a}\right|\)
\(\Rightarrow2< \left|\dfrac{a+b-c}{b}+\dfrac{b+c-a}{c}+\dfrac{a+c-b}{a}\right|< 4\)
Đặt ( a+b -c ; b+c -a ; a+c -b ) = (x ; y ;z )
=> ( a;b;c ) = \(\left(\dfrac{x+y}{2};\dfrac{y+z}{2};\dfrac{z+x}{2}\right)\)
=> Ta sẽ chứng minh
\(1< \dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}< 2\)
Ta có
\(\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}>\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}\)
\(=1\)
;;;
\(\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}=3-\left(\dfrac{y}{x+y}+\dfrac{z}{y+z}+\dfrac{x}{x+z}\right)< 3-1=2\)
=> ĐPCM
