mitrov wrote: Find the smallest number $k$, such that $ \frac{l_a+l_b}{a+b}<k$ for all triangles with sides $a$ and $b$ and bisectors $l_a$ and $l_b$ to them, respectively. Proposed by Sava Grodzev, Svetlozar Doichev, Oleg Mushkarov and Nikolai Nikolov If $a=b=1$ and $c\rightarrow2$ then $l_a=l_b\rightarrow\frac{4}{3}$ and $\frac{l_a+l_b}{a+b}\rightarrow\frac{4}{3}$. We'll prove that $\frac{l_a+l_b}{a+b}\leq\frac{4}{3}$. Let $a=y+z$, $b=x+z$ and $c=x+y$, where $x$, $y$ and $z$ are positives. Hence, $\frac{l_a+l_b}{a+b}\leq\frac{4}{3}\Leftrightarrow\frac{\frac{2bc\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}}{b+c}+\frac{2ac\sqrt{\frac{1+\frac{a^2+c^2-b^2}{2ac}}{2}}}{a+c}}{a+b}\leq\frac{4}{3}\Leftrightarrow$ $\Leftrightarrow\frac{(a+c)\sqrt{bc(a+b+c)(b+c-a)}+(b+c)\sqrt{ac(a+b+c)(a+c-b)}}{(a+b)(a+c)(b+c)}\leq\frac{4}{3}\Leftrightarrow$ $\Leftrightarrow\tfrac{(x+2y+z)\sqrt{x(x+y)(x+z)(x+y+z)}+(2x+y+z)\sqrt{y(x+y)(y+z)(x+y+z)}}{(2x+y+z)(x+2y+z)(x+y+2z)}\leq\frac{2}{3}$. By AM-GM $\sqrt{x(x+y)(x+z)(x+y+z)}\leq\frac{1}{4}(2x+z)(2x+2y+z)$ and $\sqrt{y(x+y)(y+z)(x+y+z)}\leq\frac{1}{4}(2y+z)(2x+2y+z)$. Hence, it remains to prove that $\frac{(x+2y+z)(2x+z)(2x+2y+z)+(2x+y+z)(2y+z)(2x+2y+z)}{(2x+y+z)(x+2y+z)(x+y+2z)}\leq\frac{8}{3}$, which is $4(x+y)(x-y)^2+(20x^2+44xy+20y^2+29xz+29yz+10z^2)z\geq0$, which is obvious. Id est, the answer is $\frac{4}{3}$.