Find all polynomials with degree $\leq n$ and nonnegative coefficients, such that $P(x)P(\frac{1}{x}) \leq P(1)^2$ for every positive $x$
1. Let \( P(x) = \sum_{i=0}^{j} a_i x^i \) be a polynomial of degree \( j \leq n \) with nonnegative coefficients \( a_i \).
2. Then, \( P\left(\frac{1}{x}\right) = \sum_{i=0}^{j} a_i \frac{1}{x^i} \).
3. We need to show that \( P(x)P\left(\frac{1}{x}\right) \leq P(1)^2 \) for every positive \( x \).

4. Consider the expression \( P(1) = \sum_{i=0}^{j} a_i \). Then, we have:
   \[
   P(1)^2 = \left( \sum_{i=0}^{j} a_i \right)^2
   \]

5. Using the Cauchy-Schwarz inequality, we get:
   \[
   \left( \sum_{i=0}^{j} a_i \right)^2 \leq \left( \sum_{i=0}^{j} a_i x^i \right) \left( \sum_{i=0}^{j} a_i \frac{1}{x^i} \right)
   \]

6. Therefore, we have:
   \[
   P(1)^2 \leq P(x)P\left(\frac{1}{x}\right)
   \]

7. Given that \( P(x)P\left(\frac{1}{x}\right) \leq P(1)^2 \), we combine the inequalities to get:
   \[
   P(1)^2 \leq P(x)P\left(\frac{1}{x}\right) \leq P(1)^2
   \]

8. This implies:
   \[
   P(x)P\left(\frac{1}{x}\right) = P(1)^2
   \]

9. For the equality to hold in the Cauchy-Schwarz inequality, the sequences \( \{a_i x^i\} \) and \( \{a_i \frac{1}{x^i}\} \) must be proportional. This means that \( P(x) \) must be of the form \( a_j x^j \) for some \( a_j \in \mathbb{R}^+ \).

10. Therefore, the polynomial \( P(x) \) must be of the form:
    \[
    P(x) = a_j x^j
    \]
    where \( a_j \) is a nonnegative real number.

The final answer is \( \boxed{ P(x) = a_j x^j } \) for some \( a_j \in \mathbb{R}^+ \).