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Inequality Introductory Problem 2

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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

Solution

Multiply both sides by $2$ :

$2\sum_{k=1}^{n}a_{k}^{2}\ge 2(a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1})$

By subtracting each side by the RHS, you result in:

$(a_1-a_n)^2+(a_2-a_1)^2+(a_3-a_2)^2+\cdots+(a_n-a_{n-1})^2\ge 0$

Which is always true.

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