Problem 1724

We are tasked with finding all real-coefficient polynomials $f(x)$ that satisfy the following conditions:

1. $f(x) = a_0 x^{2n} + a_2 x^{2n - 2} + \cdots + a_{2n - 2} x^2 + a_{2n}$, where $a_0 > 0$.
2. $\sum_{j=0}^n a_{2j} a_{2n - 2j} \leq \binom{2n}{n} a_0 a_{2n}$.
3. All the roots of $f(x)$ are imaginary numbers with no real part.

To solve this, we note that by condition (iii), the roots of $f(x)$ are purely imaginary. Let the roots be $\pm \alpha_1 i, \pm \alpha_2 i, \ldots, \pm \alpha_n i$, where $\alpha_j > 0$ for all $j$. This implies that $f(x)$ can be factored as:
$$f(x) = a_0 (x^2 + \alpha_1^2)(x^2 + \alpha_2^2) \cdots (x^2 + \alpha_n^2).$$

Using Vieta's formulas, we express the coefficients $a_{2k}$ in terms of the roots:
$$\frac{a_{2k}}{a_0} = \sum_{|S|=k, S \in Z'} \left( \prod_{a \in S} a \right)^2,$$
where $Z' = \{ \alpha_1 i, \alpha_2 i, \ldots, \alpha_n i \}$.

Condition (ii) can be rewritten using these coefficients:
$$\sum_{j=0}^n a_{2j} a_{2n - 2j} \leq \binom{2n}{n} a_0 a_{2n}.$$

By applying the Cauchy-Schwarz inequality and the Vandermonde identity, we find that equality holds if and only if all $\alpha_j$ are equal. Therefore, all $\alpha_j$ must be the same, say $\alpha$. Thus, the polynomial $f(x)$ simplifies to:
$$f(x) = a_0 (x^2 + \alpha^2)^n,$$
where $a_0 > 0$ and $\alpha \in \mathbb{R} \setminus \{0\}$.

Hence, the polynomials that satisfy the given conditions are:
$$f(x) = a_0 (x^2 + \alpha^2)^n,$$
where $a_0 > 0$ and $\alpha \in \mathbb{R} \setminus \{0\}$.

The answer is: $\boxed{f(x)}$ = a_0 (x^2 + $\alpha^2)^n \text{}$ where } a_0 > 0 $\text{}$ and } ${\alpha \in \mathbb{R} \setminus \{0\}}.$