Let $\alpha$ be a real number. We need to determine all polynomials $P(x)$ with real coefficients satisfying:

$$P(2x + \alpha) \leq (x^{20} + x^{19})P(x)$$

for all real numbers $x$.

### Step-by-Step Solution

1. Analyzing the inequality:

The inequality $P(2x + \alpha) \leq (x^{20} + x^{19})P(x)$ involves comparing $P(2x + \alpha)$ with the product $(x^{20} + x^{19})P(x)$.

2. **Assume non-zero polynomial $P(x)$:**

Suppose $P(x)$ is not the zero polynomial. Let $d$ be the degree of $P(x)$. Then, the degree of $P(2x + \alpha)$ is also $d$. The expression $(x^{20} + x^{19})P(x)$ is a polynomial of degree $20 + d$.

3. Leading coefficient behavior:

Notice for large values of $x$, the term $(x^{20} + x^{19})$ behaves approximately like $x^{20}$. Hence, $(x^{20}+x^{19})P(x)$ has significantly higher degree terms than $P(2x + \alpha)$ unless $d=0$ (i.e., $P(x)$ is a constant polynomial).

4. **Considering constant $P(x)$:**

For $P(x)$ constant, we take $P(x) = c$ where $c \neq 0$. Then the inequality becomes $c \leq (x^{20} + x^{19})c$. This holds for all $x$ provided $c = 0$.

5. Correctness:

If there exists even a single $x$ for which the inequality does not hold due to positive $P(x)$, then $P(x)$ cannot remain non-zero across all real $x$ because $P(x)$ can outweigh the factor of zero or negative $(x^{20} + x^{19})$.

Thus, the only polynomial $P(x)$ that satisfies the given inequality for all real numbers $x$ is the zero polynomial.

$$\boxed{P(x) \equiv 0}$$

This conclusion adheres strictly to the inequality constraint that $P(x)$ must meet for all values of $x$. Hence, $P(x) \equiv 0$ is the only suitable and valid solution.