Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$. Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$.
How can I use the first two equations to help solve the third?  I'm stuck.  Any solutions are greatly appreciated!

When $a = b = c = \frac{1}{\sqrt3}$,
we have $ab + bc + ca = 1$ and $a + b + c + abc = \frac{10\sqrt3}{9}$.
Indeed, the minimum of $a + b + c + abc$ is $\frac{10\sqrt3}{9}$.
It suffices to prove that, for all $a, b, c\in (0, 1)$ with $ab + bc + ca = 1$,
$$a + b + c + abc\ge \frac{10\sqrt3}{9}.$$
Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$.
We need to prove that
$$p + r \ge \frac{10\sqrt3}{9}. \tag{1}$$
Using $p^2 \ge 3q$, we have $p \ge \sqrt3$.
If $p \ge \frac{10\sqrt3}{9}$, clearly (1) is true.
If $\sqrt3 \le p < \frac{10\sqrt3}{9}$, using $r \ge \frac{4pq - p^3}{9}$ (3 degree Schur), we have
\begin{align*}
 p + r - \frac{10\sqrt3}{9} 
 &\ge p + \frac{4pq - p^3}{9} - \frac{10\sqrt3}{9}\\
 &= \frac{13}{9}p - \frac{1}{9}p^3
 - \left(\frac{13}{9}\sqrt3 - \frac{1}{9}(\sqrt3)^3\right)\\
 &= \frac19(p - \sqrt3)(10 - p^2 - p\sqrt 3)\\
 &\ge 0.
\end{align*}
We are done.
