Proof: Let the first term of a geometric sequence be $a$, the common ratio be $q$, and the number of terms be $n$, then the last term and the sum of this sequence are
$$a_{n}=a q^{n-1}, S=a\left(1+q+q^{2}+\cdots+q^{n-1}\right) .$$

For an arithmetic sequence with the first term $a$ and the last term $a q^{n-1}$, its sum is
$$S^{\prime}=\frac{n}{2}\left(a+a_{n}\right)=\frac{a n}{2}\left(1+q^{n-1}\right)$$

Since $q>0$, we have
$$1+q^{n-1} \geqslant q^{k}+q^{n-k-1}(k=0,1,2, \cdots, n-1)$$

Therefore, by adding all the inequalities, we get
$$n\left(1+q^{n-1}\right) \geqslant \sum_{k=0}^{n-1}\left(q^{k}+q^{n-k-1}\right)=2\left(1+q+q^{2}+\cdots+q^{n-1}\right)$$

Since $a>0$, it follows that $S<S^{\prime}$, thus the proposition is proved.