The problem statement is: Two of the altitudes of a triangle are not smaller than the corresponding sides. Find the angles of the triangle.
I first thought that it might be useful to look over these https://en.wikipedia.org/wiki/List_of_triangle_inequalities#Altitudes however I found nothing that might help. I really don't see how the fact that two altitudes are bigger than two sides links with the angles. One thing I notice is that the triangle, when attempted a representation of it, should be acute, such that the altitudes are on the projection of the sides. I think that I should also consider using either the law of sines, or the law of cosines to link the sides to the angles, but I don't know how to link all these results. I think that by applying the circumradius theorem, we obtain: $h_a=\frac{bc}{2R} \leq \frac{h_bh_c}{2R}$.
All in all, I just looked at this problem from all sides and still could not find a way to solve it or make good use of some results.
$\alpha$, $\beta$ and $\gamma$ are the angles opposite to $a$, $b$ and $c.$
Without loss of generality, we can assume $h_a \geq a$ and $h_b \geq b.$ We know that $$a\sin\gamma = h_b \\ b\sin\gamma = h_a $$ which means $$ ab \geq ab\sin^2\gamma = h_ah_b \geq ab $$ Therefore, $ab = ab\sin^2\gamma$, which in turn means $\sin^2\gamma = 1$ or $\gamma=90^{\circ}$.
Now we know that $h_a = b$ and $h_b = a$. We have $$ a = h_b \geq b = h_a \geq a $$ Which means $h_a = h_b = a = b.$ Therefore, $\alpha = \beta = 45^{\circ}$