Example 8 Find all real-coefficient polynomials $f(x)$ that satisfy the following conditions:
(1) $f(x)=a_{0} x^{2 n}+a_{1} x^{2 n-2}+\cdots+a_{2 n-2} x^{2}+a_{2 n}, a_{0}>0$;
(2) $\sum_{j=0}^{n} a_{2 j} a_{2 n-2 j} \leqslant c_{2 n}^{n} a_{0} a_{2 n}$;
(3) $f(x)$ has $2 n$ roots that are all purely imaginary.
Let $g(t)=a_{0} t^{n}-a_{2} t^{n-1}+\cdots+(-1)^{j} a_{2 j} t^{n-j}+\cdots+(-1)^{n} a_{2 n}$, then $f(x)=(-1)^{n} g\left(-x^{2}\right)$. Suppose $\pm \mathrm{i} \beta_{1}, \pm \mathrm{i} \beta_{2}, \cdots, \pm \mathrm{i} \beta_{n}$ are the $2 n$ pure imaginary roots of the polynomial $f(x)$, then the $n$ roots of the polynomial $g(t)$ are $t_{j}=\beta_{j}^{2}>0, j=1,2, \cdots, n$. Using Vieta's formulas and important inequalities to solve.

Solution First, let $g(t)=a_{0} t^{n}-a_{2} t^{n-1}+\cdots+(-1)^{j} a_{2 j} t^{n-j}+\cdots+(-1)^{n} a_{2 n}$, then
$$
f(x)=(-1)^{n} g\left(-x^{2}\right) .
$$

Suppose $\pm \mathrm{i} \beta_{1}, \pm \mathrm{i} \beta_{2}, \cdots, \pm \mathrm{i} \beta_{n}$ are the $2 n$ roots of the polynomial $f(x)$ (assuming $\beta_{j}>0, j=1,2, \cdots, n$), then the $n$ roots of the polynomial $g(t)$ are $t_{j}=\beta_{j}^{2}>0, j=1,2, \cdots, n$. By Vieta's formulas, we have
$$
\frac{a_{2 j}}{a_{0}}=\sum_{1 \leqslant k_{1} \leqslant k_{2} \leqslant \cdots \leqslant k_{j} \leqslant n} t_{k_{1}} t_{k_{2}} \cdots t_{k_{j}}, j=1,2, \cdots, n .
$$
(In the following few lines, the summation range for the symbol $\sum$ is $1 \leqslant k_{1} \leqslant k_{2} \leqslant \cdots \leqslant k_{j} \leqslant n$.) By the Cauchy-Schwarz inequality, for $j=1,2, \cdots, n$ we have
$$
\begin{aligned}
\left(C_{n}^{j}\right)^{2} \frac{a_{2 n}}{a_{0}} & \left.=\left[\sum \sqrt{t_{k_{1}} t_{k_{2}} \cdots t_{k_{j}}} \cdot \frac{\sqrt{\frac{a_{2 n}}{a_{0}}}}{\sqrt{t_{k_{1}} t_{k_{2}} \cdots t_{k_{j}}}}\right]\right]^{2} \\
& \leqslant\left(\sum t_{k_{1}} t_{k_{2}} \cdots t_{k_{j}}\right)\left[\sum \frac{\frac{a_{2 n}}{a_{0}}}{t_{k_{1}} t_{k_{2}} \cdots t_{k_{j}}}\right]=\frac{a_{2 j}}{a_{0}} \cdot \frac{a_{2 n-2 j}}{a_{0}}
\end{aligned}
$$

Noting that $C_{2 n}^{n} \frac{a_{2 n}}{a_{0}}=\sum_{j=0}^{n}\left(C_{n}^{j}\right)^{2}$, and by (1) and condition (2), we get
$$
C_{2 n}^{n} \frac{a_{2 n}}{a_{0}}=\sum_{j=0}^{n}\left(C_{n}^{j}\right)^{2} \frac{a_{2 n}}{a_{0}} \leqslant \sum_{j=0}^{n} \frac{a_{2 j}}{a_{0}} \cdot \frac{a_{2 n-2 j}}{a_{0}} \leqslant C_{2 n}^{n} \frac{a_{2 n}}{a_{0}}
$$

From (2), we see that for $j=1,2, \cdots, n$, the equality holds in (1). According to the Cauchy-Schwarz inequality and the condition for equality, we know $t_{1}=t_{2}=\cdots=t_{n}$.
Let this positive number be $r^{2}$, then $\frac{a_{2 j}}{a_{0}}=C_{n}^{j} r^{2 j}$, i.e., $a_{2 j}=a_{0} C_{n}^{j} r^{2 j}, j=1,2, \cdots, n$,
thus $f(x)=a_{0}\left(x^{2}+r^{2}\right)^{n}\left(a_{0}>0, r>0\right)$.

It is easy to verify that such a polynomial $f(x)$ satisfies all the conditions given in the problem.