Sửa đề c/m \(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
Ta có: \(\hept{\begin{cases}a< b\Rightarrow2a< a+b\\c< d\Rightarrow2c< c+d\\m< n\Rightarrow2m< m+n\end{cases}}\)
=>\(2\left(a+c+m\right)< a+b+c+d+m+n\)
=>\(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
Sửa đề: Chứng minh: \(\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
Ta có: \(\hept{\begin{cases}a< b\Rightarrow2a< a+b\\c< d\Rightarrow2c< c+d\\m< n\Rightarrow2m< m+n\end{cases}}\)
\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)