Question

[Tô Thu Huyền]

cho 6 số nguyên dương a<b<c<d<m<n

chứng minh rằng \(\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{1}{2}\)

Answer 1

[Huong San]

\(a< b=>2a< a+b\\ c< d=>2c< c+d\\ m< n=>2m< m+n\)

Suy ra \(2\left(a+c+m\right)< \left(a+b+c+d+m+n\right)\) do đó:

\(\dfrac{a+c+m}{a+b+c+d+m+n}< \dfrac{1}{2}\)