Find the $\lim_{n\to\infty}\text{inf} \left(\frac {x_0^2}{ x_1}+\frac {x_1^2}{ x_2}+\cdots \frac {x_{n-1}^2}{ x_n}\right)$ Here is my problem:

If $(x_n)_{n\in\Bbb N}$ is non-increasing, $x_0=1$ and $\lim_{n\to\infty} x_n=0$, does the following infimum exist?
$$\inf_{(x_n)_{n\in\Bbb N}} \sum_{k\in\Bbb N}\frac{x_k^2}{x_{k+1}}$$

Here is the "meat" of my attempt:

$$(a-2b)^2\geq 0$$

So we have,
$$(x_i-2x_{i+1})^2\geq 0$$
$$\frac{x_i^2}{x_{i+1}} \geq 4(x_i-x_{i+1})$$ where $i=0,1,2,\cdots ,n$
$$\frac {x_0^2}{ x_1}+\frac {x_1^2}{ x_2}+\cdots \frac {x_{n-1}^2}{ x_n}\geq 4( x_0-x_1+x_1-x_2+\cdots + x_{n-1}-x_n)=4(1-x_n)$$
Finally we get,

$$\lim_{n\to\infty}\text{inf} \left(\frac {x_0^2}{ x_1}+\frac {x_1^2}{ x_2}+\cdots \frac {x_{n-1}^2}{ x_n}\right) \geq \lim_{n\to\infty} 4(1-x_n)= 4$$
$$\lim_{n\to\infty}\text{inf} \left(\frac {x_0^2}{ x_1}+\frac {x_1^2}{ x_2}+\cdots \frac {x_{n-1}^2}{ x_n}\right)=4$$

Question 1: Is this solution correct?
Question 2 : Is it possible to find a completely different method to solve this problem?
I'm more interested in question 2. Is it possible to solve this problem with any pure calculus technique?
 A: For any $n \in \mathbb{Z}_{+}$, let $\mathcal{S}_n \subset (0,1]^n$ be the set of non-increasing finite sequences of $n$ terms taking values from $(0,1]$.
Let $f_n : \mathcal{S}_n \to \mathbb{R}$ be the function
$$\mathcal{S}_n \ni y = (y_1,y_2,\ldots,y_n)
\quad\mapsto\quad \sum_{k=1}^n \frac{y_{k-1}^2}{y_k}
\quad\text{ where }\quad y_0 = 1
$$
and $M_n = \inf_{y\in \mathcal{S}_n} f_n(y)$.
Notice for any $y = (y_1,\ldots,y_{n+1}) \in \mathcal{S}_{n+1}$, the sequence $z = \left(\frac{y_2}{y_1},\ldots,\frac{y_{n+1}}{y_1}\right) \in \mathcal{S}_n$. 
Rewrite $f_{n+1}(y)$ in terms of $y_1$ and $z$, we get
$$f_{n+1}(y) = \frac{1}{y_1} + y_1 f_n(z) \ge \frac{1}{y_1} + y_1 M_n
\stackrel{\rm AM \ge GM}{\ge} 2\sqrt{M_n}$$
Taking infimum over $y$, this leads to
$$M_{n+1} = \inf_{y\in \mathcal{S}_{n+1}} f_{n+1}(y) \ge 2\sqrt{M_n}$$
It is easy to see $M_1 = 1$. From this, we can deduce
$$M_2 \ge 2\sqrt{M_1} = 2
\implies M_3 \ge 2\sqrt{M_2} = 2^{1+\frac12}
\implies \cdots$$
In general, we have
$$M_n \ge L_n \stackrel{def}{=} 2^{1+\frac12+\frac1{2^2} + \cdots + \frac{1}{2^{n-2}}} = 2^{2-2^{2-n}}$$
This lower bound $L_n$ is achievable for any $n$. For $n = 1$, this is obvious.
Let's say for a particular $n$, the lower bound $L_n$ is achieved by sequence
$z = (z_1,z_2,\ldots,z_n) \in \mathcal{S}_n$. i.e
$f_n(z) = M_n = L_n = 2^{2-2^{2-n}}$. 
Consider the sequence $y = \left(\frac{1}{\sqrt{L_n}}, \frac{z_1}{\sqrt{L_n}},\ldots,\frac{z_n}{\sqrt{L_n}}\right)$. It is easy to see $y \in \mathcal{S}_{n+1}$. Furthermore
$$f_{n+1}(y) = \sqrt{L_n} + \frac{1}{\sqrt{L_n}} f_n(z) = 2\sqrt{L_n} = L_{n+1}$$
This means lower bound $L_{n+1}$ is achieved by $y$. By induction, the lower bound $L_n$ is achievable for all $n$. i.e.
$$M_n = \inf_{y\in \mathcal{S}_{n}} f_{n}(y) = \min_{y\in \mathcal{S}_{n}} f_{n}(y) = L_n\quad\text{ for all }\quad n > 0$$
From this, we have
$$\lim_{n\to\infty}\inf_{x\in \mathcal{S}_\infty} \left(\frac {x_0^2}{x_1}+ \cdots + \frac{x_{n-1}^2}{ x_n}\right)
= \lim_{n\to\infty}\inf_{x\in \mathcal{S}_n} f_n(x) =
 \lim_{n\to\infty} 2^{2-2^{2-n}} = 4$$
