3. Given that $a, b$ are unequal positive numbers, insert two sets of numbers
$$
\begin{array}{l}
x_{1}, x_{2}, \cdots, x_{n}, \\
y_{1}, y_{2}, \cdots, y_{n},
\end{array}
$$

such that $a, x_{1}, x_{2}, \cdots, x_{n}, b$ form an arithmetic sequence, and $a, y_{1}, y_{2}, \cdots, y_{n}, b$ form a geometric sequence. Then, among the following inequalities:
(1) $\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}$
$$
>\sqrt{a b}+\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} \text {, }
$$
(2) $\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}>\frac{a+b}{2}$,
(3) $\sqrt[n]{y_{1} y_{2} \cdots y_{n}}<\sqrt{a b}$,
(4) $\sqrt[n]{y_{1} y_{2} \cdots y_{n}}<\frac{a+b}{2}-\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}$, the true propositions are ( ).
(A) (1), (3)
(B) (1), (4)
(C) (2), (3)
(D) (2), (4)
3. (B).

Solution 1: From the arithmetic sequence, we have $x_{1}+x_{n}=x_{2}+x_{n-1}=\cdots=a+b$, thus
$$
\begin{array}{l}
\frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \\
=\frac{\left(x_{1}+x_{n}\right)+\left(x_{2}+x_{n-1}\right)+\cdots+\left(x_{n}+x_{1}\right)}{2 n} \\
=\frac{n(a+b)}{2 n}=\frac{a+b}{2}=\sqrt{a b}+2\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} . \\
>\sqrt{a b}+\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} .
\end{array}
$$

It is clear that (1) is true, (2) is false.
From the geometric sequence, we know $y_{1} y_{n}=y_{2} y_{n-1}=\cdots=a b$, thus
$$
\begin{array}{l}
\sqrt[n]{y_{1} y_{2} \cdots y_{n}}=\sqrt[2 n]{\left(y_{1} y_{n}\right)\left(y_{2} y_{n-1}\right) \cdots\left(y_{n} y_{1}\right)} \\
=\sqrt[2 n]{(a b)^{n}}=\sqrt{a b}=\frac{a+b}{2}-2\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} \\
<\frac{a+b}{2}-\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2} .
\end{array}
$$

It is clear that (3) is false, (4) is true.
In summary, (1) and (4) are true.
Solution 2: Take $a=1, b=64, n=2$, we can verify that (2) and
(3) do not hold, thus negating (A), (C), (D), and hence (B) is true.