The inequality $p<\frac{a+m_b}{b+m_a}
Let in triangle $ABC$ $BC=a$ and $AC=b$, $m_a, m_b -$ the medians drawn from the vertex $A$ and $B$. Find the greatest real number $p$ and the lowest real $q$, such that inequality
$$p<\frac{a+m_b}{b+m_a}<q$$
hold.
My attempt:
I used formula:
$$m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$

Hint for $p=\frac{1}{4}$:
We know 
\begin{align}
p<\frac{a+m_b}{b+m_a}
\end{align}
holds for all triangles. Fix the corners $A,C$ of the triangle and let $B$ to converge to $C$. Since we always have
\begin{align*}
b + \frac{a}{2} > m_a > b - \frac{a}{2} \\
\frac{b}{2} + a  > m_b > \frac{b}{2} - a
\end{align*}
when $a \rightarrow 0$ we are going to have
\begin{align*}
m_a &\rightarrow b \\
m_b &\rightarrow \frac{b}{2} \\
a &\rightarrow 0
\end{align*}
So we should have
\begin{align}
p &\leq \frac{b/2}{b+b} \\
p &\leq \frac{1}{4}
\end{align}
If $p > \frac{1}{4}$ you can find some $a,b$ such that the inequality doesn't satisfy.
Edit: Verifying $p=\frac{1}{4}$ (with some cheap inequalities)
Case-1: $b\geq m_a$
We know $m_b > \frac{b}{2} - a$, therefore
\begin{align*}
4m_b + 4a &> 2b - 4a + 4a \\
&> 2b  > b + m_a
\end{align*}
Case-2: $b < m_a$
Call the intersection of medians from $A,B$ as $G$. Define the foot of the median from $B$ as $E$. The triangle inequality in $AGE$ gives us $\frac{m_b}{3} >  \frac{2m_a}{3} - \frac{b}{2}$. Using this we have
\begin{align*}
4m_b + 4a &> 8m_a - 6b +4a \\
&> 8m_a - 6b  > m_a + b
\end{align*}
