Inequality Introductory Problem 2

Problem

Show that $\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1$ .

Solutions

First Solution

Working backwards from the next inequality we solve the origninal one: $ $\begin{array}{rcl} (\sum_{k = 1}^{n - 1} (a_{k} - a_{k + 1})^{2}) + (a_{n} - a_{1})^{2} & \geq & 0 \\ 2 \cdot (\sum_{k = 1}^{n} a_{k}^{2}) - 2 ((\sum_{k = 1}^{n - 1} a_{k} a_{k + 1}) + a_{n} a_{1}) & \geq & 0 \\ 2 \cdot (\sum_{k = 1}^{n} a_{k}^{2}) & \geq & 2 ((\sum_{k = 1}^{n - 1} a_{k} a_{k + 1}) + a_{n} a_{1}) \\ 2 \cdot \sum_{k = 1}^{n} a_{k}^{2} & \geq & 2 (a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n} + a_{n} a_{1}) \\ \sum_{k = 1}^{n} a_{k}^{2} & \geq & (a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n} + a_{n} a_{1}) \end{array}$ $ (Error compiling LaTeX. Unknown error_msg)

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Inequality Introductory Problem 2

Problem

Solutions

First Solution