The left-hand side is easy:just apply arithmetic-geometric mean inequality for the denominator and you will get it is never larger than 1: root(1+x(0)+...+x(i-1))root(x(i)+...x(n)<=(1+x(0)+...(x(n))/2=1

I think I've done the right hand side - I'll post it tomorrow because it requires a picture.

Suppose $1+x_0+x_1+...+x_i=cos{a_i}$,and then it can be proved with sum $sin x<x$.

HINT.let $\theta_i=\arc \sin (x_0+\cdots+x_i)\ ,\ i=0,1,\dots,n$ therefore $\sqrt{1+x_0+\cdots+x_{i-1}}.\sqrt{x_i+\dots+x_n}=\cos\theta_{i-1}$

Here is my solution (unfortunately, without the picture): - first, consider an unit circle. Its circumference is $2\pi$ - we will prove that the given sum is smaller than the quarter of the circumference, i.e. $\frac{\pi}{2}$. - let's consider the quarter of our circle (O is the center of the circle). By putting on the radius points $A_{1}, A_{2},...,A{n}$ such that $OA_{1}=x_{1}, A_{1}A_{2}=x_{2},...,A_{n-1}A_{n}=x_{n}$ we have our condition that the sum of x_i is 1. - let $S_{i}=x_{1}+...+x_{i-1}$. By drawing a line from $A_{i-1}$ perpendicular to the radius (let B be the point where the line touches the cirlce) we get a right-angled triangle with hypotenuse OB and two other sides OA_{i-1} and A_{i-1}B. Of course OB=1 and OA_{i-1}=S_{i}, so $A_{i-1}B=\sqrt[2]{1-S_{i}^{2}}$. - now we draw similar line from $A_{i}$, paralell to A_{i-1}B, and let C be the point where the line touches the circle. $A_{i-1}A_{i}=x_{i}$. Obviously the segment BC is smaller than the arc BC, and the sum of arcs is $\frac{\pi}{2}$, so if we prove that the segment BC is greater than $\frac{x_{i}}{\sqrt[2]{1+x_{0}+...+x_{i-1}}\sqrt[2]{x_{i}+...+x_{n}}}$, we are done (this reasoning is applied to every x_{i}) - our expression $\frac{x_{i}}{\sqrt[2]{1+x_{0}+...+x_{i-1}}\sqrt[2]{x_{i}+...+x_{n}}}$ is equivalent to $\frac{x_{i}}{\sqrt[2]{1-S_{i}^{2}}}$ (it can't be displayed properly , but it's an expression with $S_{i}$). We have to prove that this is smaller than the length of the segment BC. It requires an awful lot of calculations, and then we have to prove an inequality that I proved using calculus (derivatives etc.). I don't post it here because it's somewhat boring and definitely not elegant, but it seems to me that it's correct.

saeid and dingdongdog, could you explain your solutions in detail?

Maybe you can use the fact:$\int_0^{1} \frac 1{\sqrt{1-x^2}}=\frac{\pi}2$

let $\theta_i=\arc \sin (x_0+\cdots+x_i)\ ,\ \ \ \ i=0,1,\dots,n$ therefor$\sqrt{1+x_0+\cdots+x_{i-1}}.\sqrt{x_i+\dots+x_n}=\cos\theta_{i-1}$ now the inequality is equal to: $\sum_{i=1}^n \frac{\sin\theta_i-\sin\theta_{i-1}}{\cos\theta_{i-1}}<\frac{\pi}{2}$ notice that: $\sin\theta_i-\sin\theta_{i-1}=2.\cos\frac{\theta_i+\theta_{i-1}}{2}.\sin\frac{\theta_i-\theta_{i-1}}{2}<\cos\theta_{i-1}(\theta_i-\theta_{i-1})$ because $\theta_{i-1}<\theta_i$ and if $x>0,\sin x<x$ hence: $\sum_{i=1}^n\frac{\sin\theta_i-\sin\theta_{i-1}}{\cos\theta_{i-1}}<\sum_{i=1}^n(\theta_i-\theta_{i-1})=\theta_n-\theta_0<\frac{\pi}{2}$ Is it clear.

You surely meant $sin\theta_i=x_0+\cdots+x_i$, not $\theta_i=\arc \sin (x_0+\cdots+x_i)$?

I did not solve this all by myself, but this is certainly a good inequality. Let $y_i$ be the partial sum of the first $i$ terms in $\{x_n\}$. Then, for the lower bound, note that $$S = \sum_{i=1}^n \frac{y_i-y_{i-1}}{\sqrt{1+y_{i-1}} \cdot \sqrt{1-y_{i-1}}} = \sum_{i=1}^n \frac{y_i-y_{i-1}}{\sqrt{1-y_{i-1}^2}} \geq \sum_{i=1}^n \frac{y_i - y_{i-1}}1 = y_n = y_0 = 1.$$For the RHS bound, let $y_i = \sin \theta_i$ for some $0=\theta_0<\theta_1<\cdots<\theta_n = \frac{\pi}2$. Then, \begin{align*} S &= \sum_{i=1}^n \frac{\sin \theta_i - \sin \theta_{i-1}}{\cos \theta_{i-1}} \\ &= \sum_{i=1}^n \frac{2\cos \frac{\theta_i + \theta_{i-1}}2 \sin \frac{\theta_i-\theta_{i-1}}2}{\cos \theta_{i-1}} \\ &< \sum_{i=1}^n 2\sin \frac{\theta_i - \theta_{i-1}}2 \\ &< \sum_{i=1}^n 2 \cdot \frac{\theta_i - \theta_{i-1}}2 = \theta_n - \theta_0 =\frac{\pi}2 \end{align*}because $\cos$ is strictly decreasing on $\left[0, \frac{\pi}2\right]$ and $\sin x < x$ for any angle in that interval. We are done.