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!
 A: We can use the method of Lagrange Multipliers, which is generally useful when you want to minimize or maximize something given a constraint. We have that
$$\mathcal{L}(a,b,c,\lambda) = (a+b+c+abc)-\lambda(ab+ac+bc-1)$$
Taking partials, we get that
$$\frac{\partial \mathcal{L}}{\partial a} = 1+bc-\lambda(b+c)$$
$$\frac{\partial \mathcal{L}}{\partial b} = 1+ca-\lambda(c+a)$$
$$\frac{\partial \mathcal{L}}{\partial c} = 1+ab-\lambda(a+b)$$
Setting each to $0$, we get that
$$\lambda = \frac{1+ab}{a+b} = \frac{1+ac}{a+c} = \frac{1+bc}{b+c}$$
in addition to our original constraint. Expanding $\frac{1+ab}{a+b} = \frac{1+ac}{a+c}$ yields that
$$a+c+a^2b+abc = a+b+a^2c+abc$$
$$(a^2-1)b = (a^2-1)c$$
$$a=\pm 1\mathrm{\ or\ } b=c$$
Since $0<a<1$, we then get that $b=c$. In addition we can find that $a=b$ using similar methods, and then $a=b=c$.
We now use our constraint:
$$1=ab+ac+bc=3a^2$$
$$a=\frac{\sqrt{3}}{3}$$
and our minimum value becomes
$$\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}+\left(\frac{\sqrt{3}}{3}\right)^3$$
$$\sqrt{3}+\frac{\sqrt{3}}{9}$$
$$\frac{10\sqrt{3}}{9}$$
A: 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.
