<s>[INST] 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:
$$m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$$
 [/INST] 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!
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