Comment: Expanding the left-hand side of the desired inequality gives
$$\begin{align*} & \sum_{k=1}^{n} \left( 1 - \frac{k}{\sum_{i=1}^{n} ix_i^2} \right)^2 \cdot \frac{x_k^2}{k} \\ &= \sum_{k=1}^{n} \frac{x_k^2}{k} - \sum_{k=1}^{n} \frac{2x_k^2}{\sum_{i=1}^{n} ix_i^2} + \sum_{k=1}^{n} \frac{kx_k^2}{\left(\sum_{i=1}^{n} ix_i^2\right)^2} \\ &= \sum_{k=1}^{n} \frac{x_k^2}{k} - \frac{1}{\sum_{i=1}^{n} ix_i^2} \sum_{k=1}^{n} \frac{1}{2x_k^2} + \frac{1}{\left(\sum_{i=1}^{n} ix_i^2\right)^2} \sum_{k=1}^{n} kx_k^2 \\ &= \sum_{k=1}^{n} \frac{x_k^2}{k} - \frac{2}{\sum_{i=1}^{n} ix_i^2} + \frac{1}{\sum_{i=1}^{n} ix_i^2} \\ &= \sum_{k=1}^{n} \frac{x_k^2}{k} - \frac{1}{\sum_{i=1}^{n} ix_i^2}. \end{align*}$$
We want to show that
$$\sum_{k=1}^{n} \frac{x_k^2}{k} - \frac{1}{\sum_{i=1}^{n} ix_i^2} \le \left(\frac{n-1}{n+2}\right)^2 \sum_{k=1}^{n} \frac{x_k^2}{k} = \sum_{k=1}^{n} \frac{x_k^2}{k} - \frac{4n}{(n+1)^2} \sum_{k=1}^{n} \frac{x_k^2}{k}$$
or
$$\frac{4n}{(n+1)^2} \sum_{k=1}^{n} \frac{x_k^2}{k} \le \frac{1}{\sum_{i=1}^{n} i x_i^2},$$
that is,
$$\left( \sum_{k=1}^{n} \frac{x_k^2}{k} \right) \left( \sum_{k=1}^{n} kx_k^2 \right) \le \frac{(n+1)^2}{4n}. \qquad \textcircled{1}$$
We present two proofs of the above inequality.

Solution 1. We rewrite 1 as
$$4n \left( \sum_{k=1}^{n} \frac{x_k^2}{k} \right) \left( \sum_{k=1}^{n} kx_k^2 \right) \le (n+1)^2.$$
By the AM-GM inequality, we have
$$\begin{aligned} 4n \left( \sum_{k=1}^{n} \frac{x_k^2}{k} \right) \left( \sum_{k=1}^{n} kx_k^2 \right) &= 4 \left( \sum_{k=1}^{n} \frac{nx_k^2}{k} \right) \left( \sum_{k=1}^{n} kx_k^2 \right) \\ &\le \left( \sum_{k=1}^{n} \frac{nx_k^2}{k} + \sum_{k=1}^{n} kx_k^2 \right)^2 \\ &= \left( \sum_{k=1}^{n} \left( \frac{n}{k} + k \right) x_k^2 \right)^2. \end{aligned}$$
It suffices to show that
$$\frac{n}{k} + k \le n + 1$$
or
$$0 \le nk + k - k^2 - n = (n-k)(k-1),$$
which is evident.
Now, we consider the equality case. Note that the last inequality is strict for $1 < k < n$. Hence, we must have $x_2 = \cdots = x_{n-1} = 0$. For the AM-GM inequality to hold, we must have
$$\sum_{k=1}^{n} \frac{nx_k^2}{k} = \sum_{k=1}^{n} kx_k^2 \text{ or } nx_1^2 + x_n^2 = x_1^2 + nx_n^2$$
with $x_1^2 + x_n^2 = 1$. We must have $x_1^2 = x_n^2 = \frac{1}{2}$ and $x_2 = \cdots = x_{n-1} = 0$.

Solution 2. (By Lynnelle Ye) We write ④ as
$$(n + 1)^2 - 4n \left( \sum_{k=1}^{n} \frac{x_k^2}{k} \right) \left( \sum_{k=1}^{n} kx_k^2 \right) \ge 0.$$
Note that the left-hand side of the above inequality is the discriminant of the quadratic (in $t$):
$$f(t) = n(x_1^2 + 2x_2^2 + \cdots + nx_n^2)t^2 - (n+1)t + \left(x_1^2 + \frac{x_2^2}{2} + \cdots + \frac{x_n^2}{n}\right).$$
It suffices to show that $f(t)$ has a real root. Because the leading coefficient of $f(x)$ is $n(x_1^2 + 2x_2^2 + \cdots + nx_n^2)$, which is positive, it remains to be shown that $f(t) \le 0$ for some $t$. Because $x_1^2 + \cdots + x_n^2 = 1$, we have
$$\begin{aligned} f(t) &= n(x_1^2 + 2x_2^2 + \cdots + nx_n^2)t^2 - (n+1)(x_1^2 + x_2^2 + \cdots + x_n^2)t + \left(x_1^2 + \frac{x_2^2}{2} + \cdots + \frac{x_n^2}{n}\right)t \\ &= \sum_{k=1}^{n} \left( nkx_k^2t^2 - (n+1)x_k^2t + \frac{x_k^2}{k} \right) \\ &= \sum_{k=1}^{n} x_k^2(nkt - 1)\left(t - \frac{1}{k}\right). \end{aligned}$$
It is easy to see that $f(\frac{1}{n}) \le 0$ because for $t = \frac{1}{n}$, each summand
$$x_k^2(nkt - 1)\left(t - \frac{1}{k}\right) = \frac{x_k^2(k-1)(k-n)}{nk} \le 0,$$
completing our proof. For the equality case of the given inequality, we must have equality case for the above inequality for every $k$; that is, $x_k = 0$ for $2 \le k \le n - 1$. It is then not difficult to obtain that
$$$$
x_1^2 = x_2^2 = \frac{1}{2}. \quad \square