Ta có: `(1+2+3+...+a)/a = ([(a-1):1+1].(a+1):2)/a`
`= (a(a+1):2)/a`
`=(a+1):2`
`=a/2 + 1/2`
`(1+2+3+...+b)/b = ([(b-1):1+1].(b+1):2)/b`
`= (b(b+1):2)/b`
`=(b+1):2`
`= b/2 + 1/2`
Vì `(1+2+3+...+a)/a ` < `(1+2+3+...+b)/b`
=> `a/2 +1/2 < b/2 + 1/2`
=> `a/2 < b/2`
=> `a<b`
Vậy `a<b`