자연계열 (오후) 1번

(1-1) 주어진 부등식으로부터,$\displaystyle \begin{aligned}\frac1{a_{n+1}}\le\frac{a_n^2+n-1}{na_n}\Rightarrow\frac n{a_{n+1}}&\le a_n+\frac{n-1}{a_n}\\\Rightarrow\frac n{a_{n+1}}-\frac{n-1}{a_n}&\le a_n\end{aligned}$

(1-2)$\displaystyle \begin{aligned}a_1+a_2+\cdots+a_n&\ge\left(\frac1{a_2}-\frac0{a_1}\right)+\left(\frac2{a_3}-\frac1{a_2}\right)+\cdots+\left(\frac n{a_{n+1}}-\frac{n-1}{a_n}\right)\quad\text{((1-1)에 의해)}\\&=\frac n{a_{n+1}}\end{aligned}$

(1-3) (1) $\displaystyle n=2$일 때,$\displaystyle \begin{aligned}a_1+a_2&\ge a_1+\frac1{a_1}\quad\text{((1-1)에 의해)}\\&\ge2\quad\text{(제시문(가)에 의해)}\end{aligned}$(2) $\displaystyle a_1+\cdots+a_k\ge k$라 가정하자.(i) $\displaystyle a_{k+1}\ge1$이면 자명하다:$\displaystyle a_1+a_2+\cdots+a_k+a_{k+1}\ge k+1$

(ii) $\displaystyle a_{k+1}<1$이면,$\displaystyle \begin{aligned}a_1+\cdots+a_k+a_{k+1}&\ge\frac k{a_{k+1}}+a_{k+1}\quad\text{((1-2)에 의해)}\\&=\frac{k-1}{a_{k+1}}+\left(\frac1{a_{k+1}}+a_{k+1}\right)\\&>k-1+2=k+1\quad\text{(제시문(가)에 의해)}\end{aligned}$