# 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!

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

$$\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}+\left(\frac{\sqrt{3}}{3}\right)^3$$

$$\sqrt{3}+\frac{\sqrt{3}}{9}$$

$$\frac{10\sqrt{3}}{9}$$