Given that $a>b$. If $a+Ha\leq b+Hb$, find $c$, where $a,b,c$ are the sides $BC,AC,AB$ of triangle $ABC$ and $Ha$,$Hb$ are altitudes through $A$ and $B$ respectively.
I tried something and got that $2R \leq c$, where $R$ is circumradius. I don't know if it's correct, and if even it is....it's not sufficient. Please help me with this problem
from the given inequality we get by using $$h_a=\frac{2S}{a},h_b=\frac{2S}{b}$$ $$a-b\le 2S\left(\frac{a-b}{ab}\right)$$ or $$ab\le 2S$$ with $$S=\frac{1}{2}ab\sin(\gamma)$$ we get $$1\le \sin(\gamma)$$ can you finish?