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Given arbitrary real numbers $a_i$,

Prove that

$\sum_{i =1}^{26} \frac{a_i}{\sum_{j =0}^{i} a_j^2} \leq \sqrt{26}$

where $a_0 = 1$

So it will look like:

$\frac{a_1}{(1+a_{1}^2)} + \frac{a_2}{(1+ a_{1}^2 + a_{2}^2)} + \cdots + \frac{a_{26}}{(1+ a_1^2+ \cdots + a_{26}^2)}$

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    Sorry. Fixed it. Thanks for correcting!2012-12-21

2 Answers 2

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Here is a proof for the more general inequality $ \sum_{i=1}^n \frac{a_i}{1 + \sum_{j=1}^i a_j^2} \leq \sqrt{n}. $ where $a_1, \ldots, a_n$ range over $\mathbb{R}$. The method is a little cumbersome so I would be interested in a more direct proof.

I assume this problem comes from a math competition and would be interested to know its source.

Let $S_n$ denote the sum. Clearly we can assume that each $a_i \geq 0$, as otherwise we could increase $S_n$ by swapping the sign of one of the variables. Now make the change of variables $ a_i = t_i \sqrt{1 + a_1^2 + \cdots + a_{i-1}^2} $ for each $i$ so that $ \frac{a_i}{1 + \sum_{j=1}^{i-1} a_j^2 + a_i^2} = \frac{1}{\sqrt{1 + \sum_{j=1}^{i-1} a_j^2}} \frac{t_i}{1 + t_i^2} $ and furthermore $ \frac{1}{\sqrt{1 + \sum_{j=1}^{i-1} a_j^2}} = \prod_{j=1}^{i-1} \frac{1}{\sqrt{1 + t_j^2}}. $ Now we have $ S_n = \sum_{i=1}^n \frac{t_i}{1 + t_i^2} \prod_{j=1}^{i-1} \frac{1}{\sqrt{1 + t_j^2}}. $ Grouping terms, we get $ S_n = \frac{t_1}{1 + t_1^2} + \frac{1}{\sqrt{1 + t_1^2}} \left( \frac{t_2}{1 + t_2^2} + \frac{1}{\sqrt{1 + t_2^2}} \left( \cdots \right)\right) $ We therefore define $f : [n] \to \mathbb{R}^+$ inductively by $ f(n) = \max_{t_n \geq 0} \frac{t_n}{1 + t_n^2} $ and $ f(k) = \max_{t_k \geq 0} \frac{t_k}{1 + t_k^2} + f(k+1) \frac{1}{\sqrt{1 + t_k^2}}. $ Then we have $S_n \leq f(1)$ (and in fact, $f(1) = \max_{t_1, \ldots, t_n} S_n$).

Note that if $M$ is an upper bound for $f(k+1)$, then $ \max_{t_k \geq 0} \frac{t_k}{1 + t_k^2} + M \frac{1}{\sqrt{1 + t_k^2}} $ is an upper bound for $f(k)$. It therefore suffices to show that $ f(k) \leq \sqrt{n+1-k} $ inductively in $k$, starting at $n$.

We have the base case $ f(n) = \max_{t_n \geq 0} \frac{t_n}{1 + t_n^2} \leq \frac{1}{2}. $ It therefore suffices for us to show $ \frac{t}{1 + t^2} + \sqrt{m} \frac{1}{\sqrt{1 + t^2}} \leq \sqrt{m+1} $ for every $m \geq 1$ uniformly in $t \geq 0$. With the trivial bound $ \frac{t}{1 + t^2} \leq \frac{t}{\sqrt{1 + t^2}} $ it suffices to show $ \frac{t + \sqrt{m}}{\sqrt{1 + t^2}} \leq \sqrt{m+1}. $ Squaring, it suffices to show $ \frac{t^2 + 2 t \sqrt{m} + m}{1 + t^2} \leq m + 1 $ Or, equivalently, $ t^2 + 2 t \sqrt{m} + m \leq m + 1 + (m + 1) t^2 $ or $ 2 t \sqrt{m} \leq 1 + m t^2. $ But this is the AM-GM inequality for the pair $(1, mt^2)$, and the proof is complete.

Note that the method of proof showed that the inequality is not sharp for any $n$.

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    in general, any inequality involving a polynomial or a rational function can be deduced from finitely many “classical” inequalities by “inducting on the degree”. But would writing out all the constants make us wiser ?2012-12-23
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Suppose that $n$ is a positive integer and that $x\in \mathbb{R}^{n+1}$. By the Cauchy-Schwarz inequality, we have $\sum_{k=0}^n |x_k| \le \left(\sum_{k=0}^n x^2_k\right)^{1/2} \left(\sum_{k=0}^n 1\right)^{1/2} = \sqrt{n+1}\|x\| $ Dividing we get $\sum_{k=0}^n {|x_k|\over {\|x\|^2}} \le {\sqrt{n+1}\over \|x\|}$ Since the first coordiante of your vector is 1, you have $\|x\|\ge 1$.

From this I am able to obtain $ \sum_{k=0}^n {|x_k|\over {\|x\|^2}} \le \sqrt{n+1}$

Perhaps you can sharpen this.

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    Considering that $y$ou c$a$n get equ$a$lit$y$ in the CS, I highly doubt you can lower the upper bound... The good news is that the LHS is very different than the one in the inequality, so not being able to improve the RHS in this inequality doesn't disprove the original one....2012-12-21