# Induction Proof: Inequality involving Summation of Products with Squared Terms

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? – Thomas Andrews Mar 13 '12 at 20:29

For $a_1=1$, $a_2=2$ and $x_1=0,x_2=1$, this would mean:
$$\frac{4}{5}\leq \frac{2}{3}$$
thanks for spotting this flaw in the question, what about if $x_{i} > 0$ only. – Comic Book Guy Mar 13 '12 at 20:55
It's still false. If it's true for $(x_1,x_2)$ then it's true for $(x_1+c,x_2+c)$. But that means that $(x_1,x_2)=(1,2)$ is still wrong when $(a_1,a_2)=(1,2)$. @Hardy – Thomas Andrews Mar 13 '12 at 20:58
If $x_i=x$ for all $i$ then the expressions are always equal. The formulas on the left and right side are weighted averages of the $x_i$ with weights $a_i^2$ on the left and weights $a_i$ on the right. But the weighted average of a constant set of data is always just that constant. – Thomas Andrews Mar 13 '12 at 21:04