Example 10 Try to find all real-coefficient polynomials $f(x)$ that satisfy the following conditions:
(1) $f(x)=a_{0} x^{2 n}+a_{2} x^{2 n-2}+\cdots+a_{2 n-2} x^{2}+a_{2 n}, a_{0}>0$;
(2) $\sum_{j=0}^{n} a_{2 j} \cdot a_{2 n-2 j} \leqslant C_{2 n}^{n} a_{0} a_{2 n}$;
(3) $f(x)$ has $2 n$ roots that are all purely imaginary.
Analysis: Let $\pm \mathrm{i} \beta_{j}(1 \leqslant j \leqslant n)$ be the $2n$ roots of $f(x)$, then $f(x)=a_{0} \prod_{j=1}^{n}\left(x-\mathrm{i} \beta_{j}\right)\left(x+\mathrm{i} \beta_{j}\right)=$ $a_{0} \prod_{j=1}^{n}\left(x^{2}+\beta_{j}^{2}\right)$, and solve by comparing coefficients.

Solution: Let $\pm \mathrm{i} \beta_{j}$ be the $2n$ roots of $f(x)$ (assuming $\beta_{j}>0,1 \leqslant j \leqslant n$), then
$$
f(x)=a_{0} \prod_{j=1}^{n}\left(x-\mathrm{i} \beta_{j}\right)\left(x+\mathrm{i} \beta_{j}\right)=a_{0} \prod_{j=1}^{n}\left(x^{2}+\beta_{j}^{2}\right).
$$

By comparing the coefficients of $f(x)$ on both sides, we have $a_{2 j}=a_{0} \sum_{1 \leqslant t_{1}0(1 \leqslant j \leqslant n)$.
where $a_{2 n}=a_{0} \prod_{j=1}^{n} \beta_{j}^{2}$. Let $\pi=\prod_{j=1}^{n} \beta_{j}^{2}$, then
$$
\begin{aligned}
& C_{2 n}^{n} a_{0} a_{2 n} \geqslant \sum_{j=0}^{n} a_{2 j} \cdot a_{2 n-2 j}=\sum_{j=0}^{n-1} a_{2 j} \cdot a_{2 n-2 j}+2 a_{0} a_{2 n} \\
= & a_{0}^{2} \sum_{j=1}^{n-1}\left(\sum_{1 \leqslant t_{1}0, a_{0}>0\right)$.
It is easy to verify that such a polynomial $f(x)=a_{0}\left(x^{2}+\beta^{2}\right)^{n}\left(\beta>0, a_{0}>0\right)$ satisfies all the conditions given in the problem.