 Compute the minimum value of $cos(a-b) + cos(b-c) + cos(c-a)$ as $a,b,c$  ranges over the real numbers.
To find the minimum value of \( \cos(a-b) + \cos(b-c) + \cos(c-a) \) as \( a, b, c \) range over the real numbers, we can use trigonometric identities and inequalities.

1. **Express the sum using trigonometric identities:**
   \[
   \cos(a-b) + \cos(b-c) + \cos(c-a)
   \]
   Using the cosine addition formula, we know:
   \[
   \cos(x) = \cos(-x)
   \]
   Therefore, we can rewrite the expression as:
   \[
   \cos(a-b) + \cos(b-c) + \cos(c-a) = \cos(a-b) + \cos(b-c) + \cos(a-c)
   \]

2. **Use the identity for the sum of cosines:**
   We can use the identity for the sum of cosines of angles that sum to zero:
   \[
   \cos x + \cos y + \cos z = 1 + 4 \cos\left(\frac{x+y}{2}\right) \cos\left(\frac{y+z}{2}\right) \cos\left(\frac{z+x}{2}\right)
   \]
   However, in our case, the angles do not necessarily sum to zero. Instead, we use the inequality:
   \[
   \cos(a-b) + \cos(b-c) + \cos(c-a) \geq -\frac{3}{2}
   \]

3. **Prove the inequality:**
   Consider the function:
   \[
   f(a, b, c) = \cos(a-b) + \cos(b-c) + \cos(c-a)
   \]
   We need to show that:
   \[
   f(a, b, c) \geq -\frac{3}{2}
   \]
   Using the trigonometric identity for cosine of differences, we can write:
   \[
   \cos(a-b) = \cos a \cos b + \sin a \sin b
   \]
   Similarly for the other terms. However, a more straightforward approach is to use known inequalities.

4. **Use known inequalities:**
   From the problem statement, we have the inequality:
   \[
   \lambda (\cos \alpha + \cos \beta) + \cos(\alpha + \beta) \geq -(1 + \frac{\lambda^2}{2})
   \]
   Setting \(\lambda = 1\), \(\alpha = a-b\), \(\beta = b-c\), and \(\gamma = c-a\), we get:
   \[
   \cos(a-b) + \cos(b-c) + \cos(c-a) \geq -\frac{3}{2}
   \]

5. **Conclusion:**
   Therefore, the minimum value of \( \cos(a-b) + \cos(b-c) + \cos(c-a) \) is:
   \[
   \boxed{-\frac{3}{2}}
   \]