Hmm. Let $P(x) = \sum_{k=0}^d a_kx^k$. Then $P(x)P(\frac {1} {x}) = (\sum_{k=0}^d a_kx^k)(\sum_{k=0}^d a_kx^{-k}) \geq (\sum_{k=0}^d a_k)^2 = P(1)^2$ for all $x>0$. Given the other inequality, it follows $P(x)P(\frac {1} {x}) = P(1)^2$, which means $(\sum_{k=0}^d a_kx^k)(\sum_{k=0}^d a_{d-k}x^{k}) = P(1)^2x^d$ for all $x>0$, so the two polynomials are identical. Looking at the degrees of the monomials it follows immediately that the only possibility is $P(x) = a_dx^d$, which clearly checks out.

Suppose the $x^0$ term has non-zero coefficient. Then, as $ x \rightarrow \infty$, we may write $P(x)=cx^n+\dots+a_0$, from which it follows that $P(x)P(\frac 1 x) = O(x^n)$, which exceeds any fixed number, including $P(1)^2$, for $x$ large enough; if the first non-zero coefficient except $x^n$ is of higher order than $x^0$, the same reasoning applies (if it is $x^k$, then the product will be $O(x^{n-k})$). So there isn't any term other than $cx^n$, and this polynomial clearly works.

Say, \( P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \), where all coefficients are non-negative. Given: \[ P(x) \cdot P\left(\frac{1}{x}\right) \leq (P(1))^2 \quad \text{----(1)} \] From the Cauchy-Schwarz inequality: \[ P(x) \cdot P\left(\frac{1}{x}\right) \geq (a_n + a_{n-1} + \ldots + a_0)^2 = (P(1))^2 \quad \text{----(2)} \] Thus, we must have: \[ P(x) \cdot P\left(\frac{1}{x}\right) = (P(1))^2 \quad \text{(From equations 1 and 2)} \] Now comparing the coefficients, we get: \[ P(x) = a_k x^k, \text{ for some } a_k \in \mathbb{R}^+ \]