Do  a < b < c < d < m < n 
=> 2c < c + d 
m< n => 2m < m+ n 
=> 2c + 2a +2m = 2 ( a + c + m) < a +b + c + d + m + n) 
Do đó :
\(\dfrac{\text{(a + c + m)}}{\left(a+b+c+d+m+n\right)}\) < \(\dfrac{1}{2}\)

Do  a < b < c < d < m < n 
=> 2c < c + d 
m< n => 2m < m+ n 
=> 2c + 2a +2m = 2 ( a + c + m) < a +b + c + d + m + n) 
Do đó :
(a + c + m)/(a + b + c + d + m + n) < 1/2(đcpcm)

Đề sai cho mình sửa lại :

Cho 6 số nguyên dương a < b < c < d < m < n

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

Bài giải:

Ta có :a < b \(\Rightarrow\) 2a < a + b   ;  c < d \(\Rightarrow\) 2c < c + d  ;  m < n \(\Rightarrow\) 2m < m + n

Suy ra 2a + 2c + 2m = 2(a + c + m) < (a + b + c + d + m + n). Do đó

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

do a<b<c<d<m<n

=> a+c+m < b+d+n

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

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

Do a < b < c < d < m < n

=> a + c + m < b + d + n

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

=> a + c + m / a + b + c + d + m + n < 1/2 ( đpcm)

Do a < b < c < d < m < n

=> a + c + m < b + d + n

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

=> a + c + m / a + b + c + d + m + n < 1/2 ( đpcm)

Do a<b<c<d<m<n

\(\Rightarrow\)a+c+m<b+d+n

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

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

k cho mình nha

Do a < b < c < d < m < n

=> a + c + m < b + d + n

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

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

Ta có :

a < b \(\Rightarrow\)  2a < a + b

c < d \(\Rightarrow\)  2c < c + d

m < n \(\Rightarrow\)   2m < m + n

Suy ra :

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

Do đó :

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

a < b ⇒ 2a < a + b ; c < d ⇒ 2c < c + d ; m < n ⇒ 2m < m + n 

Suy ra 2a + 2c + 2m = 2(a + c + m) < (a + b + c + d + m + n). Do đó 

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