Find the following limit $$\lim_{n\to+\infty}\int_{x\in[0,1]^n}\frac{x_1^2 + x_2^2 + ... + x_n^2}{x_1 + x_2 + ... + x_n}dx$$.
I used Cauchy-Schwarz to find a lower bound $\frac12$: Note that
$$(x_1^2 + x_2^2 + ... + x_n^2)(1 + 1 + ... + 1) \ge (x_1 + x_2 + ... + x_n)^2$$
$$\therefore \int_{x\in[0,1]^n}\frac{x_1^2 + x_2^2 + ... + x_n^2}{x_1 + x_2 + ... + x_n}dx \ge \frac1n\int_{x\in[0,1]^n}(x_1 + x_2 + ... + x_n)dx = \frac1n(\int_{x\in[0,1]^n}x_1dx + \int_{x\in[0,1]^n}x_2dx + ... + \int_{x\in[0,1]^n}x_ndx)$$
Also
$$\int_{x\in[0,1]^n}x_idx = \frac12$$
$$\therefore \int_{x\in[0,1]^n}\frac{x_1^2 + x_2^2 + ... + x_n^2}{x_1 + x_2 + ... + x_n}dx\ge \frac12, \forall n$$
By assuming the order of all $x_i$, we can get an upper bound 1.
$$\frac{x_1^2 + x_2^2 + ... + x_n^2}{x_1 + x_2 + ... + x_n}\le \max\{\frac{x_1^2}{x_1}, \frac{x_2^2}{x_2}, ..., \frac{x_n^2}{x_n}\} = \max\{x\}$$
By symmetricity
$$\int_{x\in[0,1]^n}\frac{x_1^2 + x_2^2 + ... + x_n^2}{x_1 + x_2 + ... + x_n}dx = n\int_{x\in[0,1]^n\\\max x = x_1}\frac{x_1^2 + x_2^2 + ... + x_n^2}{x_1 + x_2 + ... + x_n}dx \\\le n\iint_{x_1\in[0,1]\\x_i\in[0,x_1]}x_1dx = n\int_{x_1\in[0,1]}x_1^ndx_1 = \frac{n}{n+1}$$
whose limit is 1.
But what's the exact limit?
Seems like this problem has something to do with expectation and variance, but I have no idea on how to do it.
