Let $T$ be a triangle with vertex $A$, $B$ and $C$ and sides lengths $a$, $b$ and $c$ opposite to their vertex. We have $c\geq b\geq a$. Consider a line segment joining midpoint of $AB$ with $C$. Let the midpoint be $L$ and length of $CL$ be $l$. Can we have an lower-bound $\alpha>0$ on $c-l$ i.e. $c-l \geq \alpha$ ?
I tried applying triangle inequality. We have $a+\frac{c}{2}> l $ which in turn implies $c-l>\frac{c}{2}-a$. However if $a \geq \frac{c}{2}$ this result is obvious. We also have $b>l$. Thus, $c-l >c-b$. If $c=b$ then again $\alpha=0$. I want an $\alpha >0$.
We can prove that $$c-l\geq\left(1-\frac{\sqrt3}{2}\right)c$$ or $$\frac{\sqrt3}{2}c\geq\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$$ or $$3c^2\geq2a^2+2b^2-c^2$$ or $$2c^2-a^2-b^2\geq0.$$ The equality occurs for $a=b=c$,
which says that $1-\frac{\sqrt3}{2}$ is a maximal value of $k$ for which the inequality $$c-l\geq kc$$ is true for all triangle whith $c\geq b\geq c$.
By the way, for $c\rightarrow0^+$ in $c-l\geq\alpha$ we obtain $0\geq\alpha$, which gives $\alpha=0$ is a maximal $\alpha$,
for which your inequality is true.