Problem 1678

1. Define the function and its properties:
Let $f(x) = \frac{x}{\sqrt{1-x}}$. We need to show that this function is convex. To do this, we compute the second derivative of $f(x)$.

2. Compute the first derivative:
$$f'(x) = \frac{d}{dx} \left( \frac{x}{\sqrt{1-x}} \right) = \frac{\sqrt{1-x} - \frac{x}{2\sqrt{1-x}}}{1-x} = \frac{(1-x) - \frac{x}{2}}{(1-x)^{3/2}} = \frac{2(1-x) - x}{2(1-x)^{3/2}} = \frac{2 - 3x}{2(1-x)^{3/2}}$$

3. Compute the second derivative:
$$f''(x) = \frac{d}{dx} \left( \frac{2 - 3x}{2(1-x)^{3/2}} \right)$$
Using the quotient rule:
$$f''(x) = \frac{(2 - 3x)'(1-x)^{3/2} - (2 - 3x)((1-x)^{3/2})'}{(1-x)^3}$$
Simplifying each term:
$$(2 - 3x)' = -3$$
$$((1-x)^{3/2})' = \frac{3}{2}(1-x)^{1/2}(-1) = -\frac{3}{2}(1-x)^{1/2}$$
Therefore:
$$f''(x) = \frac{-3(1-x)^{3/2} - (2 - 3x)(-\frac{3}{2}(1-x)^{1/2})}{(1-x)^3} = \frac{-3(1-x)^{3/2} + \frac{3}{2}(2 - 3x)(1-x)^{1/2}}{(1-x)^3}$$
Simplifying further:
$$f''(x) = \frac{-3(1-x) + \frac{3}{2}(2 - 3x)}{(1-x)^{5/2}} = \frac{-3(1-x) + 3 - \frac{9x}{2}}{(1-x)^{5/2}} = \frac{-3 + 3x + 3 - \frac{9x}{2}}{(1-x)^{5/2}} = \frac{3x - \frac{9x}{2}}{(1-x)^{5/2}} = \frac{6x - 9x}{2(1-x)^{5/2}} = \frac{-3x}{2(1-x)^{5/2}}$$
Since $x \in (0, 1)$, $f''(x) > 0$. Thus, $f(x)$ is convex.

4. Apply Jensen's Inequality:
Since $f(x)$ is convex and $\sum_{i=1}^n x_i = 1$, by Jensen's Inequality:
$$\sum_{i=1}^n f(x_i) \geq n f\left( \frac{\sum_{i=1}^n x_i}{n} \right) = n f\left( \frac{1}{n} \right)$$
Calculate $f\left( \frac{1}{n} \right)$:
$$f\left( \frac{1}{n} \right) = \frac{\frac{1}{n}}{\sqrt{1 - \frac{1}{n}}} = \frac{1}{n \sqrt{\frac{n-1}{n}}} = \frac{1}{\sqrt{n(n-1)}}$$
Therefore:
$$\sum_{i=1}^n \frac{x_i}{\sqrt{1-x_i}} \geq \frac{n}{\sqrt{n(n-1)}} = \frac{\sqrt{n}}{\sqrt{n-1}}$$

5. Prove the final inequality using Cauchy-Schwarz:
We need to show:
$$\frac{\sqrt{n}}{\sqrt{n-1}} \geq \frac{\sum_{i=1}^n \sqrt{x_i}}{\sqrt{n-1}}$$
This simplifies to:
$$\sqrt{n} \geq \sum_{i=1}^n \sqrt{x_i}$$
By the Cauchy-Schwarz inequality:
$$\left( \sum_{i=1}^n \sqrt{x_i} \right)^2 \leq n \sum_{i=1}^n x_i = n$$
Taking the square root of both sides:
$$\sum_{i=1}^n \sqrt{x_i} \leq \sqrt{n}$$
Thus:
$$\frac{\sqrt{n}}{\sqrt{n-1}} \geq \frac{\sum_{i=1}^n \sqrt{x_i}}{\sqrt{n-1}}$$

The final answer is $\boxed{ \sum_{i=1}^n \frac{x_i}{\sqrt{1-x_i}} \geq \frac{\sum_{i=1}^n \sqrt{x_i}}{\sqrt{n-1}} }$