Let $a,b,c\ge 0$ be three real numbers such that $$ab+bc+ca+2abc=1.$$ Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 2$ and determine equality cases.
Given the condition \( ab + bc + ca + 2abc = 1 \), we need to prove that \( \sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2 \) and determine the equality cases.

1. **Substitute \( a = \tan^2 A \), \( b = \tan^2 B \), and \( c = \tan^2 C \) where \( A, B, C \) are angles of a triangle:**

   Since \( a, b, c \ge 0 \), we can assume \( a = \tan^2 A \), \( b = \tan^2 B \), and \( c = \tan^2 C \) where \( A, B, C \) are angles of a triangle. This substitution is valid because the tangent function is non-negative for angles in the first quadrant.

2. **Use the identity for the sum of tangents:**

   We know that for any triangle, the identity \( \tan A \tan B + \tan B \tan C + \tan C \tan A = 1 \) holds. This can be rewritten as:
   \[
   \tan^2 A \tan^2 B + \tan^2 B \tan^2 C + \tan^2 C \tan^2 A + 2 \tan^2 A \tan^2 B \tan^2 C = 1
   \]
   which matches our given condition \( ab + bc + ca + 2abc = 1 \).

3. **Apply the AM-GM inequality:**

   To prove \( \sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2 \), we use the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   \sqrt{a} + \sqrt{b} + \sqrt{c} \ge 3 \sqrt[3]{\sqrt{a} \sqrt{b} \sqrt{c}}
   \]
   Since \( \sqrt{a} \sqrt{b} \sqrt{c} = \sqrt{abc} \), we have:
   \[
   \sqrt{a} + \sqrt{b} + \sqrt{c} \ge 3 \sqrt[3]{\sqrt{abc}}
   \]

4. **Analyze the equality case:**

   For the equality case in AM-GM, we need \( \sqrt{a} = \sqrt{b} = \sqrt{c} \). This implies \( a = b = c \). Substituting \( a = b = c \) into the given condition \( ab + bc + ca + 2abc = 1 \), we get:
   \[
   3a^2 + 2a^3 = 1
   \]
   Solving for \( a \), we find:
   \[
   a(2a^2 + 3a - 1) = 0
   \]
   Since \( a \ge 0 \), we discard \( a = 0 \) and solve the quadratic equation \( 2a^2 + 3a - 1 = 0 \):
   \[
   a = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}
   \]
   Since \( a \ge 0 \), we take the positive root:
   \[
   a = \frac{-3 + \sqrt{17}}{4}
   \]
   This value of \( a \) satisfies the given condition, and thus \( \sqrt{a} + \sqrt{b} + \sqrt{c} = 3\sqrt{a} \).

5. **Verify the inequality:**

   Since \( \sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2 \) holds for \( a = b = c \), and the equality case is when \( a = b = c = \frac{-3 + \sqrt{17}}{4} \), we conclude that the inequality is true.

\(\blacksquare\)

The final answer is \( \boxed{ \sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2 } \)