Inequality Introductory Problem 2: Difference between revisions

Latest revision as of 14:18, 23 May 2009

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

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.

@@ Line 5: / Line 5: @@
 == Solutions ==
-===First Solution===
+===Solution===
-Working backwards from the next inequality we solve the origninal one:
+Multiply both sides by <math>2</math>:
-<math>
+<center>
-\begin{eqnarray*}
+<math>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})</math>
-\left(\sum_{k=1}^{n-1} (a_k-a_{k+1})^2\right) + (a_n-a_1)^2&\ge& 0\\
+</center>
-\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\\
-\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)\\
+By subtracting each side by the RHS, you result in:
-\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)
+<center>
-\end{eqnarray*}
+<math>(a_1-a_n)^2+(a_2-a_1)^2+(a_3-a_2)^2+\cdots+(a_n-a_{n-1})^2\ge 0</math>
-</math>
+</center>
+Which is always true.