Consider the triangle which has the smallest circumcircle, called it $\triangle{ABC}$. Denote its circumcenter by $O$ and note that the circumradius is equal to $\frac{a}{2sin(\angle{A})}=\frac{b}{2sin(\angle{B})}$ Assume that $\angle{A}\leq \angle{B}\leq \angle{C}$, we have $\angle{A},\angle{B} \in (0,\frac{\pi}{2})$ Note that $\angle{AOC}=2\angle{B},\angle{BOC}=2\angle{A}$. The circumradii of $\triangle{BOC}$ and $\triangle{AOC}$ are $\frac{a}{2sin(2\angle{A})}$ and $\frac{b}{2sin(2\angle{B})}$ respectively. So $\frac{a}{2sin(\angle{A})} \leq \frac{a}{2sin(2\angle{A})}$ where $sin(2\angle{A}),sin(\angle{A})>0$. We get $sin(\angle{A})\geq sin(2\angle{A})\Rightarrow \frac{1}{2}\geq cos(\angle{A})$. So $\angle{A} \geq \frac{\pi}{3}$, similarly $\angle{B} \geq \frac{\pi}{3}$. This mean that $\angle{C}\leq \frac{\pi}{3}$ give us $\angle{C}$ is an acute angle and $sin(2c)>0,\angle{AOB}=2\angle{C}$ Thus the similar relation must hold, which give us $\angle{C}\geq \frac{\pi}{3}$. So the equality case must occur, we have $\angle{A}=\angle{B}=\angle{C}=\frac{\pi}{3}$. In another word, $\triangle{ABC}$ must be an equilateral triangle. Let $D$ denote circumcenter of $\triangle{BOC}$ We get that $\triangle{BOD}$ is a marked triangle whose smaller circumcircle, contradiction with the definition of $\triangle{ABC}$.

Just saw that this was Problem C3 here.