6. Determine the greatest real number $p$ and the smallest real number $q$, for which the inequalities

$$
p<\frac{a+t_{b}}{b+t_{a}}<q
$$

hold in any triangle $ABC$ with sides $a, b$ and medians $t_{a}, t_{b}$.

(Pavel Novotný)
Solution. We will show that the sought numbers are $p=1 / 4$ and $q=4$. It is sufficient to justify that $q=4$ (then $p=1 / 4$, since swapping sides $a, b$ changes the value of the fraction to its reciprocal).

According to the triangle inequalities, we have

$$
\frac{1}{2} a0$, then for all sufficiently small positive $b$ simultaneously we have

$$
\sqrt{4-b^{2}}>2-\varepsilon, \quad 2 b\frac{4-\varepsilon}{1+2 \varepsilon}
$$

and it is easy to choose $\varepsilon>0$ so that the last fraction is greater than any previously chosen $q$ less than 4. It is sufficient that

$$
\varepsilon<\frac{4-q}{1+2 q}
$$