Question

cho 6 số nguyên dương  a<b<c<d<m<n

CM:\(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)

Answer 1

Ta có:

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

=> \(\frac{2\left(a+c+m\right)}{a+b+c+d+m+n}< 1\)

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

Answer 2

Theo giải thiết đề bài ta có : : \(a< b< c< d< m< n\Rightarrow2a< a+b;2c< c+d;2m< m+n\)

\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{2\left(a+c+m\right)}{a+b+c+d+m+n}< \frac{\frac{a+b+c+d+m+n}{2}}{a+b+c+d+m+n}=\frac{1}{2}\)

Vậy \(\frac{a+c+m}{a+c+d+m+n}< \frac{1}{2}\) (đpcm)