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I was trying to solve one of the bounty questions (yes i know it is very ambitious for a newbie like me:-) ). But regardless of my analysis being correct or incorrect, another problem originated from my answer which required proving or disproving $\dfrac {\sum _{i=1}^{i=n}a_{i}^{2}x_{i}} {\sum _{i=1}^{i=n}a_{i}^{2}}\leq\dfrac {\sum _{i=1}^{i=n}a_{i}x_{i}} {\sum _{i=1}^{i=n}a_{i}}$ where $\forall \;a_{i}, \; a_{i} > 0$ and $\forall x_{i}$, $x_{i} \in \mathbb{R} $ and $x_{i} > 0$.

I was hoping for a proof by induction but that seems somewhat harder than imagined. I checked the n=1 case which was fine, then i assumed the inequality holds for n and started with the n+1 case and observed that $\dfrac {\sum _{i=1}^{i=n}a_{i}^{2}x_{i}} {\sum _{i=1}^{i=n+1}a_{i}^{2}}\leq\dfrac {\sum _{i=1}^{i=n}a_{i}^{2}x_{i}} {\sum _{i=1}^{i=n}a_{i}^{2}}\leq\dfrac {\sum _{i=1}^{i=n}a_{i}x_{i}} {\sum _{i=1}^{i=n}a_{i}}$

$\dfrac {\sum _{i=1}^{i=n+1}a_{i}^{2}x_{i}} {\sum _{i=1}^{i=n+1}a_{i}^{2}}\leq\dfrac {\sum _{i=1}^{i=n}a_{i}^{2}x_{i}} {\sum _{i=1}^{i=n}a_{i}^{2}} + \dfrac {a_{n+1}^{2}x_{n+1}} {a_{n+1}^{2}}\leq\dfrac {\sum _{i=1}^{i=n}a_{i}x_{i}} {\sum _{i=1}^{i=n}a_{i}}+\dfrac {a_{n+1}^{2}x_{n+1}} {a_{n+1}^{2}}$

I am unsure how to proceed from here. Any thoughts about how to approach this one would be much appreciated.

PS: Yes I'd be happy to share the bounty prize if my answer was granted the bounty along with any credit or virtual bragging rights, although how we may accomplish that may need some thought or input from you.

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    If $y_i=-x_i$ wouldn't that reverse your inequality?2012-03-13

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For $a_1=1$, $a_2=2$ and $x_1=0,x_2=1$, this would mean:

$\frac{4}{5}\leq \frac{2}{3}$

Which isn't true.

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    Thanks very much for yoru help. If you are curious about the original problem then pls checkout http://math.stackexchange.com/questions/117337/inequality-for-expected-value/119774#1197742012-03-13