Given a triangle with side lengths $a,\ b,\ c$. Let $a,\ b,\ c$ vary, find the range of $\frac{a^2+b^2+c^2}{ab+bc+ca}$.

[i]2010 Tokyo Institute of Technology Admission Office entrance exam, Problem I-2/Science[/i]
1. **Using the identity involving the sum of squares:**
   Consider the identity:
   \[
   \sum (a - b)^2 \geq 0
   \]
   Expanding this, we get:
   \[
   (a - b)^2 + (b - c)^2 + (c - a)^2 \geq 0
   \]
   Simplifying, we have:
   \[
   a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 \geq 0
   \]
   Combining like terms, we get:
   \[
   2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca \geq 0
   \]
   Dividing the entire inequality by 2, we obtain:
   \[
   a^2 + b^2 + c^2 \geq ab + bc + ca
   \]
   Therefore:
   \[
   \frac{a^2 + b^2 + c^2}{ab + bc + ca} \geq 1
   \]

2. **Using the triangle inequality:**
   The triangle inequality states that for any triangle with sides \(a\), \(b\), and \(c\):
   \[
   a + b > c, \quad b + c > a, \quad \text{and} \quad c + a > b
   \]
   Multiplying each inequality by the appropriate side, we get:
   \[
   a(b + c) > a^2, \quad b(c + a) > b^2, \quad \text{and} \quad c(a + b) > c^2
   \]
   Adding these inequalities, we have:
   \[
   a(b + c) + b(c + a) + c(a + b) > a^2 + b^2 + c^2
   \]
   Simplifying the left-hand side, we get:
   \[
   ab + ac + bc + ba + ca + cb > a^2 + b^2 + c^2
   \]
   Which simplifies to:
   \[
   2(ab + bc + ca) > a^2 + b^2 + c^2
   \]
   Dividing both sides by \(ab + bc + ca\), we obtain:
   \[
   2 > \frac{a^2 + b^2 + c^2}{ab + bc + ca}
   \]
   Therefore:
   \[
   \frac{a^2 + b^2 + c^2}{ab + bc + ca} < 2
   \]

3. **Combining the results:**
   From the two derived inequalities, we have:
   \[
   1 \leq \frac{a^2 + b^2 + c^2}{ab + bc + ca} < 2
   \]

The final answer is \(\boxed{1 \leq \frac{a^2 + b^2 + c^2}{ab + bc + ca} < 2}\)