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
$$\lim_{n\to\infty}\inf_{x\in \mathcal{S}_\infty} \left(\frac {x_0^2}{x_1}+ \cdots + \frac{x_{n-1}^2}{ x_n}\right)
= \lim_{n\to\infty}\inf_{x\in \mathcal{S}_n} f_n(x) =
\lim_{n\to\infty} 2^{2-2^{2-n}} = 4$$
Using the AG-GM :
If $x_1,x_2,\cdots,x_n$ are positive real numbers, then
$$n\sqrt[n]{x_1^2x_2^2\cdots x_n^2}\leq x_1^2+x_2^2+ \cdots x_n^2$$
You can write:
$$
\int_1^\infty \cdots \int_1^\infty
\frac{dx_1 \cdots dx_n}{(x_1^{2}+\cdots + x_n^{2})^a}
\leq \frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
\frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}}
$$
But:
$$
\frac{1}{n} \int_{1}^\infty \cdots \int_{1}^\infty
\frac{dx_1 \cdots dx_n}{x_1^{2a/n}\cdots x_n^{2a/n}} =
\frac{1}{n} \int_{1}^\infty \frac{dx_1}{x_1^{2a/n}} \cdots \int_{1}^\infty \frac{dx_n}{x_n^{2a/n}}
$$
And, since $\frac{2a}{n}> 1$,for each $i \in [1,\infty)$:
$$
\int_{1}^\infty \frac{dx_i}{x_i^{2a/n}} = \left[ \frac{x_i^{1-2a/n}}{1-2a/n} \right]_1^\infty = \frac{1}{1-2a/n}
$$
Best Answer
By symmetry (i.e. by exploiting the fact that our integral is twice the integral over the sub-region $0\leq y\leq x\leq 1$) we just have to compute:
$$ I=2 \int_{0}^{1}\int_{0}^{1}\frac{x}{1+x^3 y(1+y)}\,dy\,dx = 2 \int_{0}^{1}\int_{0}^{2}\frac{x}{(1+x^3y)\sqrt{1+4y}}\,dy\,dx$$ Integrating with respect to $x$ first,
$$\begin{eqnarray*} I &=& 2\int_{0}^{2}\frac{2 \sqrt{3}\arctan\left(\frac{\sqrt{3}}{1-2y^{1/3}}\right)-2\log\left(1+y^{1/3}\right)+\log\left(1-y^{1/3}+y^{2/3}\right)}{6 y^{2/3}\sqrt{1+4y}}\,dy\\&=&\int_{0}^{2^{1/3}}\frac{2\sqrt{3}\arctan\left(\frac{\sqrt{3}}{1-2y}\right)-3\log(1+y)+\log(1+y^3)}{\sqrt{1+4y^3}}\,dy\end{eqnarray*}$$ but the resulting integrals in just one variable do not look so appealing.
Am I missing some crucial simplification that follows from replacing $y$ with a Jacobi elliptic function (maybe $\text{dn}$) or with the Weierstrass elliptic function $\wp(z)$ (corresponding to $g_2=0,g_3=-1$) then exploiting some weird/mystical product formulas?