Find all the polynomials $P(x)$ of a degree $\leq n$ with real non-negative coefficients such that $P(x) \cdot P(\frac{1}{x}) \leq [P(1)]^2$ , $ \forall x>0$.
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 \( a_n \geq 0 \).