# 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!

## 1 Answer

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

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

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

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