How to prove $\sum\limits_{i =1}^{26} \frac{a_i}{\sum_{j =0}^{i} a_j^2} \leq \sqrt{26}$ 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:
$$\frac{a_1}{(1+a_{1}^2)} + \frac{a_2}{(1+ a_{1}^2 + a_{2}^2)} + \cdots + \frac{a_{26}}{(1+ a_1^2+ \cdots + a_{26}^2)}$$
 A: 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 $n$.
A: 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.
