在 Rajendra Bhatia 的 Matrix Analysis 中, Exercise I.5.8 说: Prove that for any matrices $A,B$ we have $$\bex |\per (AB)|^2\leq \per (AA^*)\cdot \per (B^*B).
设 $\sed{a_k}_{k=1}^n$ 为等差数列, 则 $$\bex a_1+\cdots+a_n=\frac{n(a_1+a_n)}{2}. \eex$$ Ref. [Proof Without Words: Partial Sums of an Arithmetic Sequence, The College Mathematics Journal].