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

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

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

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)

Vì a < b

    c < d

     m < n

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

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

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

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

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

bạn tham khảo câu hỏi tương tự nhé!

Câu hỏi tương tự      

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

\(\Leftrightarrow2a+2b+2c< a+b+c+d+m+n\)

\(\Leftrightarrow2\left(a+b+c\right)< a+b+c+d+m+n\)

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

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

Ta có :

a < b \(\Rightarrow\)2a < a + b \(\Rightarrow\)\(\frac{a}{a+b}< \frac{1}{2}\)

c < d \(\Rightarrow\)2c < c + d \(\Rightarrow\)\(\frac{c}{c+d}< \frac{1}{2}\)

m < n \(\Rightarrow\)2m < m + n \(\Rightarrow\)\(\frac{m}{m+n}< \frac{1}{2}\)

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

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

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

\(a< b\Rightarrow2a< a+b\)

\(c< d\Rightarrow2c< c+d\)

\(m< n\Rightarrow2m< m+n\)

\(\Rightarrow2a+2c+2m< a+b+c+d+m+n\)

\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)

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

Ta có:

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

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

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

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)

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 )