Find the least real number $K$ such that for all real numbers $x$ and $y$, we have $(1 + 20 x^2)(1 + 19 y^2) \ge K xy$.
1. **Applying the AM-GM Inequality:**
   We start by applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the terms \(1 + 20x^2\) and \(1 + 19y^2\).

   For \(1 + 20x^2\):
   \[
   1 + 20x^2 \geq 2\sqrt{20x^2} = 2\sqrt{20}|x| = 2\sqrt{20}x
   \]

   For \(1 + 19y^2\):
   \[
   1 + 19y^2 \geq 2\sqrt{19y^2} = 2\sqrt{19}|y| = 2\sqrt{19}y
   \]

2. **Combining the Inequalities:**
   Multiplying the two inequalities obtained from AM-GM:
   \[
   (1 + 20x^2)(1 + 19y^2) \geq (2\sqrt{20}|x|)(2\sqrt{19}|y|) = 4\sqrt{380}|xy|
   \]

3. **Finding the Least \(K\):**
   We need to find the least \(K\) such that:
   \[
   (1 + 20x^2)(1 + 19y^2) \geq Kxy
   \]
   From the inequality derived using AM-GM, we have:
   \[
   (1 + 20x^2)(1 + 19y^2) \geq 4\sqrt{380}|xy|
   \]
   Therefore, the least \(K\) is:
   \[
   K = 4\sqrt{380} = 8\sqrt{95}
   \]

4. **Verification with Specific Values:**
   To verify that this \(K\) is indeed the least possible, consider the values \(x = -\frac{1}{\sqrt{20}}\) and \(y = \frac{1}{\sqrt{19}}\):
   \[
   1 + 20\left(-\frac{1}{\sqrt{20}}\right)^2 = 1 + 1 = 2
   \]
   \[
   1 + 19\left(\frac{1}{\sqrt{19}}\right)^2 = 1 + 1 = 2
   \]
   \[
   (1 + 20x^2)(1 + 19y^2) = 2 \cdot 2 = 4
   \]
   \[
   Kxy = 8\sqrt{95} \left(-\frac{1}{\sqrt{20}}\right) \left(\frac{1}{\sqrt{19}}\right) = -\frac{8\sqrt{95}}{\sqrt{380}} = -8
   \]
   Since \(4 \geq -8\), the value \(K = 8\sqrt{95}\) is valid and least.

\(\blacksquare\)

The final answer is \( \boxed{ K = 8\sqrt{95} } \)