Some minor points, but no proof yet on the main point, (a): The desired sequence in (b) is $x_k=\frac1{2^k}.$ I conjecture, but can't prove, that it is the unique answer to that question. For a geometric sequence, $x_k=r^k,$ we have the following: $\sum_{k=1}^{\infty}\frac{x_{k-1}^2}{x_k}=\frac1{r(1-r)}.$ Within that special family, the $r=\frac12$ case is clearly optimal. Perhaps this problem could have been placed in the inequalities section.

========================================== AUXILIARY RESULT: The LHS in (a) is greater or equal to $4-4x_n$. Proof by induction on $n$: The case $n=1$ becomes $1/x_1 \ge 4-4x_1$, which is equivalent to $(2x_1-1)^2 \ge0$. The inductive step is $LHS(n+1)=LHS(n)+ (x_{n-1})^2/x_n \ge 4-4x_{n-1} + (x_{n-1})^2/x_n \ge 4-4x_n$. Here the first inequality follows from the inductive assumption, and the second inequality is equivalent to $(2x_n-x_{n-1})^2 \ge0$. ========================================== Now distinguish two cases: (1) First, assume that $x_n$ eventually goes below $0.001$. Then the AUXILIARY RESULT yields the desired inequality for (a). (2) Secondly, assume that $x_n$ always remains at or above $0.001$. Then every term in the LHS is at least $0.001$, and the sum of the first $4000$ terms is at least $4$.

It can be proved by induction. Consider function: $f(b_k)=a_k*b_k^{-r_k}+\frac{b_k^2}{x_k}\ge a_{k+1}b_{k+1}^{-r_{k+1}},$ were $b_{k+1}=x_k,r_{k+1}=\frac{r_k}{2+r_k},a_{k+1}=(\frac{r_k}{2}a_k)^{1-r_{k+1}}(1+\frac{2}{r_k}).$ We have $r_2=1,b_2=x_1,a_2=1, r_k=\frac{1}{2^{k-1}-1}, a_k$ - increase, and $\lim_{k \to \infty} a_k=4.$

A slightly different solution: By Cauchy, the LHS is at least: $\frac{(x_{0}+...+x_{n-1})^{2}}{x_{1}+...+x_{n}}.$ It is clear that $x_{0}=1$ and $x_{n}\le \frac{x_{1}+...+x_{n-1}}{n-1}$ We thus have that the LHS is at least: $(\frac{1+2(x_{1}+...+x_{n-1})+(x_{1}+...+x_{n-1})^{2}}{x_{1}+...+x_{n-1}})(\frac{n-1}{n})$ Applying 2-variable AM-GM to $1$ and $(x_{1}+...+x_{n-1})^{2}$ the LHS is at least $4(\frac{n-1}{n})$ We then simply choose $n$ such that $4(\frac{n-1}{n}) \ge 3.999$. Part b) is likely to occur only if both the AM-GM and Cauchy equality conditions are close to met, or if $1=x_{1}+x_{2}+...$ and $\frac{x_{0}}{x_{1}}=\frac{x_{1}}{x_{2}}=...$, respectively. This points to the geometric series with common ratio $\frac{1}{2}$ as the equality case.

b) is immediate by taking $x_i = 2^{-i}$. For a), we prove $LHS \ge 2^{2^0 + 2^{-1}+...+ 2^{2-n}}$, and taking $n\to \infty$ will yield the result. (Here $n=1$ gives the empty sum, so $LHS \ge 2^0=1$, which is trivial). The proof is not hard. Let $s_i$ be the sum of the first $i$ terms, so $s_n = s_{n-2} + \dfrac{x_{n-2}^2}{x_{n-1}} + \dfrac{x_{n-1}^2}{x_n} \ge s_{n-2} + 2^{2^0} x_{n-2} \sqrt{\dfrac{x_{n-1}}{x_n}} \ge s_{n-2} + 2^{2^0} x_{n-2}$. This in turn is $s_{n-3} + \dfrac{x_{n-3}^2}{x_{n-2}} + 2^{2^0} x_{n-2}\ge s_{n-3} + 2^{2^0 + 2^{-1}} x_{n-3}$. By repeatedly applying AM-GM to the last two terms of the sum, collecting powers of two, and cancelling variables, we eventually obtain $s_0 + \dfrac{x_0^2}{x_1} + 2^{2^0 + 2^1 + ... + 2^{3-n}}x_1 \ge 0 + 2^{2^0 + 2^1 + ... + 2^{2-n}}x_0$, as desired.

a) If $\{x_n\}$ is lower bounded by a positive real, then the result is clear. Suppose that $\{x_n\}$ isn't lower bounded py a positive real. $$\frac{1}{x_1}+\sum_{i=1}^{n-1}{\frac{x_i^2}{x_{i+1}}}=\frac{1}{x_1}+\sum_{i=1}^{n-1}\left({\frac{(x_i-2x_{i+1})^2}{x_{i+1}}+4(x_i-x_{i+1})}\right) \ge\frac{1}{x_1}+\sum_{i=1}^{n-1}{4(x_i-x_{i+1})} =\frac{1}{x_i}+ 4(x_1-x_n) \ge 4-4x_n$$Thus for a big enough $n$ we will have that the sum above is bigger than $3.999$, the inequality $(x_i-2x_{i+1})^2 \ge 0$,suggests us that $x_i=2^{-i}$ works for b), the verification is immediate.

$b)$ $x_k=\frac{1}{2^k}$. $a)$ Let $a=3.999$. Note that $\frac{x_0^2}{x_1}+ \frac{x_1^2}{x_2}+ \ldots+ \frac{x_{n-1}^2}{x_n} \ge x_0+x_1+\ldots +x_{n-1} \ge nx_n$ so if there is $k$ such that $x_k \ge \frac{a}{k}$, we are done. Now assume that $x_k< \frac{a}{k},~\forall~k \in \mathbb{N}$. By weighted AM-GM, we have $$\frac{x_0^2}{x_1}+ \frac{x_1^2}{x_2}+ \ldots+ \frac{x_{n-1}^2}{x_n} \ge (2^n-1) \sqrt[2^n-1]{\left(\frac{1}{2^{n-1}} \cdot \frac{x_0^2}{x_1}\right)^{2^{n-1}} \cdot \left( \frac{1}{2^{n-2}} \cdot \frac{x_1^2}{x_2} \right)^{2^{n-2}} \cdot \ldots \cdot \left( \frac{1}{2} \cdot \frac{x_{n-2}^2}{x_{n-1}} \right) \cdot \frac{x_{n-1}^2}{x_n}}=$$$$=(2^n-1) \sqrt[2^n-1]{\left(\frac{1}{2} \right)^{(n-1)2^{n-1}+(n-2)2^{n-2}+ \ldots + 1 \cdot 2} \cdot \frac{(x_0)^{2^n}}{x_n}}.$$

The sum $(n-1)2^{n-1}+(n-2)2^{n-2}+ \ldots 1 \cdot 2$ can be calculated in a standard way

and it equals to $(n-2)2^n+2$. Thus $$\frac{x_0^2}{x_1}+ \frac{x_1^2}{x_2}+ \ldots+ \frac{x_{n-1}^2}{x_n} > (2^n-1) \sqrt[2^n-1]{\left(\frac{1}{2} \right)^{(n-2)2^n+2} \cdot \frac{n}{a}}.$$ The RHS can be written as $$4 \left(2^{-\frac{n}{2^n-1}}- 2^{-\frac{n \cdot 2^n}{2^n-1}}\right) \cdot n^{\frac{1}{2^n-1}} \cdot \frac{1}{a^{\frac{1}{2^n-1}}}$$and its limit as $n \to \infty$ is $4(1-0) \cdot 1 \cdot 1=4,$ so we are done. The limit $\lim_{n \to \infty}n^{\frac{1}{2^n-1}}$ is $1$ because $1 \le n^\frac{1}{2^n-1} \le (2^n-1)^\frac{1}{2^n-1}$ and $n^\frac{1}{n} \to 1$ from d'Alembert criterion. The other limits are obvious.

We present two solutions. Let the expression be $S$. Approach 1 (C-S) Note that $$\left(\frac{x_0^2}{x_1} + \frac{x_1^2}{x_2} + \cdots + \frac{x_{m-1}}{x_m}\right) (x_1+x_2+\cdots+x_m) \geq (x_0+x_1+\cdots+x_{m-1})^2.$$Set $A = x_1+x_2+\cdots + x_{m-1}$, so $$S \geq \frac{(A+1)^2}{A+x_m} \geq \frac{4A}{A+x_m}$$by AM-GM. This is greater than $4-\varepsilon$ for arbitrarily small $\varepsilon$ if $x_m \to 0$. If the sequence does not approach 0, it will become arbitrarily large and the inequality will be trivially true. Approach 2 (AM-GM). Notice that \begin{align*} \frac{x_0^2}{x_1} + \frac{x_1^2}{x_2}+\cdots+\frac{x_{m-1}^2}{x_m} &\geq \frac{x_0^2}{x_1}+\cdots+\frac{x_{m-2}^2}{x_{m-1}}+x_{m-1} \\ &\geq \frac{x_0^2}{x_1} +\cdots + \frac{x_{m-3}^2}{x_{m-2}} + 2x_{m-2} \\ &\geq \cdots \\ &\geq 2^{1+\frac 12 + \frac 14 + \cdots} x_0 \to 4. \end{align*} For the second part, simply take $x_n = \frac 1{2^n}$.

a) Notice that by Cauchy-Schwartz in Engel form we have that $\sum_{i=1}^n\frac{x_{n-1}^2}{x_n}\ge\frac{(\sum_{i=0}^{n-1}x_i)^2}{\sum_{i=1}^nx_i}$ Thus this implies that we need to prove that: $\Longrightarrow (\sum_{i=0}^{n-1}x_i)^2\ge3.999\sum_{i=1}^nx_i\Longrightarrow\sum_{i=0}^{n-1}x_i\ge\sqrt{3.999\sum_{i=1}^nx_i}$ $\Longrightarrow \sum_{i=0}^{n-1}x_i\ge\frac{3.999+\sum_{i=1}^nx_i}{2}$ $\Longrightarrow 2\sum_{i=0}^{n-1}x_i\ge 1.999+\sum_{i=1}^nx_i$ $\Longrightarrow \sum_{i=0}^{n-1}x_i\ge 1.999+x_n$ Thus we only need to prove that: $\sum_{i=1}^{n-2}x_i\ge1.999$ Now let us consider $\lim_{n\to\infty}\sum_{i=1}^{n-2}x_i=\sum_{i=1}^{\infty}x_i$ by the root test for divergence we have that: $L=\lim_{n\to\infty}\sqrt[n]{x_n}$ however $x_n<1$ we have that the nth root of the limit is greater than 1, which implies that the series diverges b) Trivial to see that $x_i=\frac{1}{2^i}$ satisfies the relation. And this finishes our proof! So we are done!