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