Let $\Delta ABC$

$$\begin{gathered} BC = a \\ AC = b \\ AB = c \end{gathered}$$

and median of $\Delta ABC$

$A D=l$

$B E=m$

$A D$ is median,

Similarly, $m<\frac{a+b}{2}$ and

$\Rightarrow \frac{l+m+n}{a+b+c}<1$  ... ($\mathrm{i}$)

Also in $\Delta BGC,BG+GC>BC$

$\therefore \frac{2}{3}(m+n)>a$

Similarly, $\frac{2}{3}(n+l)>b$ and $\frac{2}{3}(m+l)>c$

$\Rightarrow \frac{l+m+n}{a+b+c}>\frac{3}{4}$   ... ($\mathrm{ii}$)

From Eqs. ($\mathrm{i}$) and ($\mathrm{ii}$), we get

$\frac{l+m+n}{a+b+c}\mathit{\in }\left(\frac{3}{4},1\right)$