已知等差数列$\{a_n\}$中$a_n>0$,求证:$\left(1+\dfrac 1{a_1}\right)\left(1+\dfrac 1{a_2}\right)\cdots \left(1+\dfrac{1}{a_n}\right)\leqslant \left(1+\dfrac{a_1+a_n}{2a_1a_n}\right)^n$.
证明 考虑倒序相乘,有\[\begin{split} \left(1+\dfrac 1{a_k}\right)\left(1+\dfrac 1{a_{n+1-k}}\right)&=1+\dfrac{1}{a_k}+\dfrac{1}{a_{n+1-k}}+\dfrac{1}{a_ka_{n+1-k}}\\&=1+\dfrac{a_1+a_n+1}{a_ka_{n+1-k}}\\&\leqslant 1+\dfrac{a_1+a_n+1}{a_1a_n}\\ &\leqslant 1+\dfrac{1}{a_1}+\dfrac{1}{a_n}+\dfrac 14\left(\dfrac 1{a_1}+\dfrac 1{a_n}\right)^2\\ &=\left[1+\dfrac 12\left(\dfrac{1}{a_1}+\dfrac{1}{a_n}\right)\right]^2\\ &=\left(1+\dfrac{a_1+a_n}{2a_1a_n}\right)^2,\end{split}\]在上式中分别令$k=1,2,\cdots ,n$,然后将各式相乘即证得原不等式成立.