The inequality $p<\frac{a+m_b}{b+m_a}<q$

Question

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: $$m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$

Answer 1

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*}

Answer 2

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;

2. $4y+3z-x<0$.

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!