1. **Assume the polynomial $P(x)$ has degree $n$ and can be written as:**
$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$
where $a_i \geq 0$ for all $i$.

2. **Consider the product $P(x) \cdot P\left(\frac{1}{x}\right)$:**
$$P\left(\frac{1}{x}\right) = a_n \left(\frac{1}{x}\right)^n + a_{n-1} \left(\frac{1}{x}\right)^{n-1} + \cdots + a_1 \left(\frac{1}{x}\right) + a_0$$
$$P\left(\frac{1}{x}\right) = a_n x^{-n} + a_{n-1} x^{-(n-1)} + \cdots + a_1 x^{-1} + a_0$$

3. **Multiply $P(x)$ and $P\left(\frac{1}{x}\right)$:**
$$P(x) \cdot P\left(\frac{1}{x}\right) = \left(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0\right) \left(a_n x^{-n} + a_{n-1} x^{-(n-1)} + \cdots + a_1 x^{-1} + a_0\right)$$

4. **Analyze the leading term as $x \to \infty$:**
- The leading term of $P(x)$ is $a_n x^n$.
- The leading term of $P\left(\frac{1}{x}\right)$ is $a_n x^{-n}$.
- Therefore, the leading term of $P(x) \cdot P\left(\frac{1}{x}\right)$ is $a_n^2$.

5. **Consider the behavior of $P(x) \cdot P\left(\frac{1}{x}\right)$ as $x \to \infty$:**
- If $a_0 \neq 0$, then $P(x) \cdot P\left(\frac{1}{x}\right)$ will have terms that grow without bound as $x \to \infty$.
- Specifically, the term $a_0 \cdot a_0$ will dominate, leading to $P(x) \cdot P\left(\frac{1}{x}\right) \to \infty$.

6. **To satisfy $P(x) \cdot P\left(\frac{1}{x}\right) \leq [P(1)]^2$ for all $x > 0$, the polynomial must be of the form:**
$$P(x) = a_n x^n$$
where $a_n$ is a non-negative real number.

7. Verify that this form satisfies the given inequality:
$$P(x) = a_n x^n \implies P\left(\frac{1}{x}\right) = a_n x^{-n}$$
$$P(x) \cdot P\left(\frac{1}{x}\right) = (a_n x^n) \cdot (a_n x^{-n}) = a_n^2$$
$$[P(1)]^2 = (a_n \cdot 1^n)^2 = a_n^2$$
$$\therefore P(x) \cdot P\left(\frac{1}{x}\right) = a_n^2 = [P(1)]^2$$

Thus, the only polynomials that satisfy the given condition are of the form $P(x) = a_n x^n$ where $a_n \geq 0$.

The final answer is $\boxed{ P(x) = a_n x^n }$ where $a_n \geq 0$.