$x_1,x_2,...,x_n$ are positive real numbers, $\sum_{i=1}^n x_i^2 = 1$, to find the minimum value of: $$\sum_{i=1}^n \frac{x_i^5}{s - x_i}$$

with $$s = \sum_{i=1}^{n}x_i$$

I used Cauchy Schwarz inequality:

$$(\sum_{i=1}^n \frac{x_i^5}{s - x_i})(\sum_{i=1}^{n}\frac{s-x_i}{x_i}) \geq 1$$

$$(\sum_{i=1}^n \frac{x_i^5}{s - x_i})\geq \frac1{\sum_{i=1}^{n}\frac{s-x_i}{x_i} }$$

the denominator is $$s \sum_{i=1}^n \frac1{x_i} - n$$

i don't know a good way to limit it above
Hint:  Using CS and Power Mean inequalities 
$$LHS \cdot \sum x_i(s-x_i)\ge \left(\sum x_i^3\right)^2\ge \frac1n$$
Further $\sum x_i(s-x_i)=s^2-1\le n-1$, so LHS $\ge \dfrac1{n(n-1)}$, with equality when all the $x_i=\frac1{\sqrt n}$.
Ok, let's use the LM

$$F(x_1,...,x_n,\lambda) = (x_1 + ... + x_n)(\frac{1}{x_1} + ... + \frac{1}{x_n}) - \lambda(x_1^2 + ... + x_n^2 - 1)$$

we must have $$\nabla F = \vec 0$$

$$\frac{\partial F}{\partial x_i} = (\frac{1}{x_1} + ... + \frac{1}{x_n}) + (x_1 + ... + x_n)\frac{-1}{x_i^2} - 2\lambda x_i = 0$$

I am convinced that this is not a good aproach