Ta có:x<y
=>x+x<y+x
\(\Rightarrow\frac{2a}{m}< \frac{a+b}{m}\)
=>2a<a+b
Mà \(x=\frac{a}{m}=\frac{2a}{2m}\)
\(y=\frac{b}{m}=\frac{2b}{2m}\)
Theo giả thuyết trên:
=>2a<a+b<2b
\(\Rightarrow\frac{2a}{2m}< \frac{a+b}{2m}< \frac{2b}{2m}\)
\(\Rightarrow x< z< y\left(DPCM\right)\)