Determine the least real number $k$ such that the inequality
$$\left(\frac{2a}{a-b}\right)^2+\left(\frac{2b}{b-c}\right)^2+\left(\frac{2c}{c-a}\right)^2+k \geq 4\left(\frac{2a}{a-b}+\frac{2b}{b-c}+\frac{2c}{c-a}\right)$$
holds for all real numbers $a,b,c$.

[i]Proposed by Mohammad Jafari[/i]
1. **Claim and Verification**:
   We claim that the least real number \( k \) such that the inequality holds for all real numbers \( a, b, c \) is \( \boxed{k = 8} \). To verify, we check if the equality can be achieved when \( (a, b, c) = (4, 1, -2) \).

2. **Substitution and Simplification**:
   Let \( u = \frac{2a}{a-b} \), \( v = \frac{2b}{b-c} \), and \( w = \frac{2c}{c-a} \). Note that \( (u-2)a = ub \), so \( \prod_{\text{cyc}}(u-2) = \prod_{\text{cyc}} u \) or \( abc = 0 \).

3. **Case Analysis**:
   - If \( abc = 0 \), without loss of generality, let \( a = 0 \). The original inequality becomes:
     \[
     \left(\frac{b}{b-c} - 1\right)^2 \geq 0
     \]
     which is evidently true.

   - If \( abc \neq 0 \), the original inequality is equivalent to finding the smallest \( k \) such that:
     \[
     f(u, v, w) = u^2 + v^2 + w^2 - 4(u + v + w) + k \geq 0
     \]
     subject to:
     \[
     g(u, v, w) := \sum uv - 4\sum u + 8 = 0
     \]

4. **Function Analysis**:
   Note that:
   \[
   u^2 + v^2 + w^2 - 4u - 4v - 4w + 8 = \sum (u-2)^2 - 4 \geq 0
   \]
   for all \( u, v, w \) such that one of \( |u|, |v|, |w| \geq 10^{100} \). Therefore, we only consider \( (u, v, w) \in [-10^{100}, 10^{100}]^3 \), which is compact. Thus, local minima of \( f \) must exist.

5. **Lagrange Multipliers**:
   To find the local minima, we use the method of Lagrange multipliers. We need to check the points \( x = (u, v, w) \) such that \( f'(x) = \lambda g'(x) \) for some real number \( \lambda \).

6. **Partial Derivatives**:
   Considering partial derivatives of \( f \) and \( g \) with respect to \( u \):
   \[
   u - 2 = \lambda (-v - w + 2)
   \]
   Similarly, differentiating with respect to \( v \) and \( w \) gives:
   \[
   (u - v) = 0 \quad \text{if} \quad \lambda \neq 1
   \]
   This implies \( u = v = w \), leading to:
   \[
   u^3 = \prod u = \prod (u-2) = (u-2)^3 \implies u = u-2
   \]
   which is a contradiction. Therefore, \( \lambda = 1 \).

7. **Summation and Simplification**:
   Adding the expressions:
   \[
   u - w = \lambda (-v - w + 2)
   \]
   and symmetric expressions, we obtain:
   \[
   \sum u = \frac{6(1 + \lambda)}{1 + 2\lambda} = 4
   \]
   Hence:
   \[
   g(u, v, w) = 0 \implies \sum uv = 4
   \]
   Therefore:
   \[
   f(x) = (\sum u)^2 - 2\sum uv - 4\sum u + k = 16 - 2 \cdot 4 - 4 \cdot 4 + 8 = 0
   \]
   Thus, \( \min f = 0 \), as desired. \(\blacksquare\)