Inequality Introductory Problem 2: Difference between revisions
New page: == Problem == Show that <math>\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1</math>. == Solutions == ===First Solution=== Working backwards from the next inequality we... |
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Working backwards from the next inequality we solve the origninal one: | Working backwards from the next inequality we solve the origninal one: | ||
<math> | <math> | ||
\begin{eqnarray*} | \begin{eqnarray*} | ||
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\end{eqnarray*} | \end{eqnarray*} | ||
</math> | </math> | ||
Revision as of 11:37, 18 May 2009
Problem
Show that 
.
Solutions
First Solution
Working backwards from the next inequality we solve the origninal one: $\begin{eqnarray*} \left(\sum_{k=1}^{n-1} (a_k-a_{k+1})^2\right) + (a_n-a_1)^2&\ge& 0\\ 2\cdot \left(\sum_{k=1}^n a_k^2\right) - 2\left(\left(\sum_{k=1}^{n-1} a_ka_{k+1}\right) +a_na_1\right)&\ge& 0\\ 2\cdot \left(\sum_{k=1}^n a_k^2\right) &\ge& 2\left(\left(\sum_{k=1}^{n-1} a_ka_{k+1}\right) +a_na_1\right)\\ 2\cdot \sum_{k=1}^n a_k^2 &\ge& 2(a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1)\\ \sum_{k=1}^n a_k^2 &\ge& (a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1) \end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)
