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:
$$m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$
 A: If $a\rightarrow0^+$ and $b=c=1$ so $\frac{a+m_b}{b+m_a}\rightarrow\frac{1}{4}$.
We'll prove that $\frac{a+m_b}{b+m_a}>\frac{1}{4}$.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$\frac{2(y+z)+\sqrt{4y(x+y+z)+(x-z)^2}}{2(x+z)+\sqrt{4x(x+y+z)+(y-z)^2}}>\frac{1}{4}$$ or
$$8y+6z-2x+4\sqrt{4y(x+y+z)+(x-z)^2}>\sqrt{4x(x+y+z)+(y-z)^2}$$ 
1. $4y+3z-x\geq0$.
After squaring of the both sides we'll obtain something obvious;


*$4y+3z-x<0$.


We need to prove that 
$$4\sqrt{4y(x+y+z)+(x-z)^2}>\sqrt{4x(x+y+z)+(y-z)^2}+2(x-4y-3z)$$
In this case we need to square twice and we'll obtain something obvious.
Id est, $p=\frac{1}{4}$.
By the same way we can obtain $q=4$.
Done!
A: 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*}
