Solution:
Wlog assume $a \geq b \geq c$. Therefore $a + b + c \geq 3c$. Now for a proof by contradiction, assume $a \leq \frac{17}{10}$ and $b \leq \frac{17}{10}$. Since $\left(\frac{17}{10}\right)^2 = \frac{289}{100} < 3$, it follows that $\frac{17}{10} < \sqrt{3}$ and therefore $ab \leq 3$. However this implies:

$$3c \leq a + b + c = abc < 3c$$

which is a contradiction.

$1 = \frac{a + b + c}{abc} = \frac{1}{bc} + \frac{1}{ca} + \frac{1}{ab}$
As all three terms are positive, at least one must be less than or equal to $1/3$ and, without loss of generality, we can assume $1/bc \leq 1/3$. But then $bc \geq 3$ and at least one of $b$ or $c$ must be greater than $\sqrt{3}$. Since $\sqrt{3} > 17/10$ we're done.

By the AM-GM inequality we have $\frac{a + b + c}{3} \geq \sqrt[3]{abc}$ and thus $abc \geq 3\sqrt[3]{abc}$, which rearranges to give $\sqrt[3]{abc} \geq \sqrt{3}$. Now wlog $a \geq b \geq c$ and so

$$a = \sqrt[3]{a^3} \geq \sqrt[3]{abc} \geq \sqrt{3} > \frac{17}{10}.$$