每日一题[619]条条大路通罗马

已知$a_n=\dfrac{3^n}{3^n+2}$，求证：$a_1+a_2+\cdots +a_n>\dfrac{n^2}{n+1}$．

证明　分析通项　分析\[\begin{split} &a_n>\dfrac{n^2}{n+1}-\dfrac{(n-1)^2}{n}\\&\Leftarrow \dfrac{3^n}{3^n+2}>\dfrac{n^3-(n-1)^2(n+1)}{n(n+1)}\\ &\Leftarrow 3^n>2n^2+2n-2\\ &\Leftarrow n\geqslant 3,\end{split} \]因此有当$n\geqslant 3$时，有\[\begin{split} &a_1+a_2+\cdots a_n\\>&\dfrac 35+\dfrac 9{11}+\left(\dfrac {3^2}4-\dfrac {2^2}3\right)+\cdots +\left[\dfrac{n^2}{n+1}-\dfrac{(n-1)^2}{n}\right]\\ =&\dfrac{n^2}{n+1}-\dfrac 43+\dfrac 35+\dfrac 9{11}\\ =&\dfrac{n^2}{n+1}+\dfrac{14}{165}>\dfrac{n^2}{n+1},\end{split} \]而当$n=1,2$时命题显然成立，因此原命题得证．

等比放缩　由于$a_n=1-\dfrac{2}{3^n+2}$，而$\dfrac{n^2}{n+1}=n-\dfrac{n}{n+1}$，因此原不等式等价于$$\dfrac{2}{3+2}+\dfrac{2}{9+2}+\dfrac{2}{27+2}+\cdots+\dfrac{2}{3^n+2}<\dfrac{n}{n+1}.$$当$n=1,2$时命题显然成立，当$n\geqslant 3$时，有\[\begin{split} &\dfrac{2}{3+2}+\dfrac{2}{9+2}+\dfrac{2}{27+2}+\cdots+\dfrac{2}{3^n+2}\\<& \dfrac 25+\dfrac 2{11}+\dfrac{2}{27}+\cdots +\dfrac{2}{3^n}\\ <&\dfrac 25+\dfrac{2}{11}+\dfrac{\dfrac{2}{27}}{1-\dfrac 13}\\<&\dfrac 34\leqslant \dfrac{n}{n+1},\end{split} \]因此原命题得证．

柯西不等式放缩　由柯西不等式$$\left(a_1+a_2+\cdots+a_n\right)\left(\dfrac 1{a_1}+\dfrac{1}{a_2}+\cdots +\dfrac{1}{a_n}\right)\geqslant n^2,$$而$$\dfrac 1{a_1}+\dfrac{1}{a_2}+\cdots +\dfrac{1}{a_n}=n+2\left(\dfrac 13+\dfrac 19+\cdots +\dfrac{1}{3^n}\right)<n+1,$$因此$a_1+a_2+\cdots +a_n>\dfrac {n^2}{n+1}$，原命题得证．