设 $m,n\in\mathbb N^{\ast}$，$m>n$，证明：$[m,n]+[m+1,n+1]>\dfrac{2mn}{\sqrt{m-n}}$．
解析 设 $m-n=k$，则\[\begin{split} LHS&=\dfrac{mn}{(m,n)}+\dfrac{(m+1)(n+1)}{(m+1,n+1)}\\ &=\dfrac{mn}{(k,n)}+\dfrac{(m+1)(n+1)}{(k,n+1)}\\ &=mn\left(\dfrac1{(k,n)}+\dfrac1{(k,n+1)}\right)+\dfrac{m+n+1}{(k,n+1)},\end{split}\]由于 $(n,n+1)=1$，于是 $(k,n)\cdot (k,n+1)\leqslant k$，结合均值不等式可得\[\dfrac1{(k,n)}+\dfrac1{(k,n+1)}\geqslant \dfrac{2mn}{\sqrt k},\]命题得证．