/bump, anyone?

Can somebody explain problem to me? I don't understand what are we given, lines or segments?

The problem is equivalent to proving that at least one pair of inequalities from these three pairs are true: $$b+h_a\geqslant c,c+h_a\geqslant b$$$$c+h_b\geqslant a,a+h_b\geqslant c$$$$a+h_c\geqslant b,b+h_c\geqslant a$$where $h_a,h_b,h_c$ are the lengths of the altitudes from the three vertices.

If $\triangle ABC$ is obtuse, then the side opposite of the obtuse angle is the longest one (let's name that side $a$). Let the other two sides of the triangle be $b$ and $c$ such that $b \ge c$. The vertex opposite of the side $c$, i.e. vertex $C$, is in fact such a vertex, as if $s$ is a cevian drawn from $C$ to $c$ (with endpoints $C$ and $S$), then we have $a + b > s$ (based on the fact that in $\triangle CSB$ $\angle S > \angle A > \angle B$, i.e.that $a > s$), $a + s > b$ (based on the fact that $a > b$), as well as $b + s > a$ (based on the fact that $s > b$ -- this can be proven the same way $a > s$ was proved--, so we have that $b + s > 2b > a$, which holds because if $a \ge 2b$ then $a >= 2b \ge b + c$, which contradicts the triangle inequality). If $\triangle ABC$ is acute, and $a \ge b \ge c$, then vertex $C$ is yet again the desired vertex, as if $s$ is a cevian from that vertex then $a + b > s$ (because $a > s$, which can be shown similarly as it was shown in the previous case), $a + s > b$ (because $a > b$), and finally, in order to prove that $b + s > a$, it is sufficient to show that $b + h > a$, where $h$ is the altitude from $C$ to $c$ (and since $s \ge h$), and if we let $SB = k$, then $b + h > a$ is equivalent to proving $b + h > \sqrt{h^2 + k^2}$ which is equivalent to $b^2 + 2bh + h^2 > h^2 + k^2$, i.e. we have to prove that $b^2 + 2bh > k^2$, which holds because $b^2 \ge c^2 > k^2$.