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 thấy:
\(\hept{\begin{cases}a< b\\c< d\\m< n\end{cases}\Rightarrow a+c+m< b+d+n}\)
\(\Rightarrow\left(a+c+m\right)+\left(a+c+m\right)< \left(a+c+m\right)+\left(b+d+n\right)\)
\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)
hay \(a+b+c+d+m+n>2\left(a+c+m\right)\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{a+c+m}{2\left(a+c+m\right)}\) ( do các tử và các mẫu đều dương )
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\) ( đpcm )