, 2021

Solution 1.
Lemma 1.1. If $a, b, c$ are non-negative numbers such that $a b-c^{2} \geqslant 1$, then
$$(a+2 b)^{2} \geqslant(b+2 c)^{2}+6$$
Proof. $(a+2 b)^{2}-(b+2 c)^{2}=(a-b)^{2}+2(b-c)^{2}+6\left(a b-c^{2}\right) \geqslant 6$.

Lemma 1.2. $\sqrt{1}+\cdots+\sqrt{n}>\frac{2}{3} n^{3 / 2}$.
Proof. Bernoulli's inequality $(1+t)^{3 / 2}>1+\frac{3}{2} t$ for $0>t \geqslant-1$ (or, alternatively, a straightforward check) gives
$$\begin{equation*} (k-1)^{3 / 2}=k^{3 / 2}\left(1-\frac{1}{k}\right)^{3 / 2}>k^{3 / 2}\left(1-\frac{3}{2 k}\right)=k^{3 / 2}-\frac{3}{2} \sqrt{k} \tag{*} \end{equation*}$$
Summing up ( $*$ ) over $k=1,2, \ldots, n$ yields
$$0>n^{3 / 2}-\frac{3}{2}(\sqrt{1}+\cdots+\sqrt{n}) .$$
Now put $y_{i}:=2 x_{i}+x_{i+1}$ for $i=0,1, \ldots, n$. We get $y_{0} \geqslant 0$ and $y_{i}^{2} \geqslant y_{i-1}^{2}+6$ for $i=1,2, \ldots, n$ by Lemma 1.1. Thus, an easy induction on $i$ gives $y_{i} \geqslant \sqrt{6 i}$. Using this estimate and Lemma 1.2 we get
$$3\left(x_{0}+\ldots+x_{n+1}\right) \geqslant y_{1}+\ldots+y_{n} \geqslant \sqrt{6}(\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n})>\sqrt{6} \cdot \frac{2}{3} n^{3 / 2}=3\left(\frac{2 n}{3}\right)^{3 / 2}$$

Solution 2.
Say that an index $i \in\{0,1, \ldots, n+1\}$ is $\operatorname{good}$, if $x_{i} \geqslant \sqrt{\frac{2}{3}} i$, otherwise call the index $i$ bad.

Lemma 2.1. There are no two consecutive bad indices.
Proof. Assume the contrary and consider two bad indices $j, j+1$ with minimal possible $j$. Since 0 is good, we get $j>0$, thus by minimality $j-1$ is a good index and we have
$$\frac{2}{3} \sqrt{j(j+1)}>x_{j} x_{j+1} \geqslant x_{j-1}^{2}+1 \geqslant \frac{2}{3}(j-1)+1=\frac{2}{3} \cdot \frac{j+(j+1)}{2}$$
that contradicts the AM-GM inequality for numbers $j$ and $j+1$.

Lemma 2.2. If an index $j \leqslant n-1$ is good, then
$$x_{j+1}+x_{j+2} \geqslant \sqrt{\frac{2}{3}}(\sqrt{j+1}+\sqrt{j+2})$$
Proof. We have
$$x_{j+1}+x_{j+2} \geqslant 2 \sqrt{x_{j+1} x_{j+2}} \geqslant 2 \sqrt{x_{j}^{2}+1} \geqslant 2 \sqrt{\frac{2}{3} j+1} \geqslant \sqrt{\frac{2}{3} j+\frac{2}{3}}+\sqrt{\frac{2}{3} j+\frac{4}{3}},$$
the last inequality follows from concavity of the square root function, or, alternatively, from the AM-QM inequality for the numbers $\sqrt{\frac{2}{3} j+\frac{2}{3}}$ and $\sqrt{\frac{2}{3} j+\frac{4}{3}}$.

Let $S_{i}=x_{1}+\ldots+x_{i}$ and $T_{i}=\sqrt{\frac{2}{3}}(\sqrt{1}+\ldots+\sqrt{i})$.

Lemma 2.3. If an index $i$ is good, then $S_{i} \geqslant T_{i}$.
Proof. Induction on $i$. The base case $i=0$ is clear. Assume that the claim holds for good indices less than $i$ and prove it for a good index $i>0$.
If $i-1$ is good, then by the inductive hypothesis we get $S_{i}=S_{i-1}+x_{i} \geqslant T_{i-1}+\sqrt{\frac{2}{3}} i=T_{i}$.
If $i-1$ is bad, then $i>1$, and $i-2$ is good by Lemma 2.1. Then using Lemma 2.2 and the inductive hypothesis we get
$$S_{i}=S_{i-2}+x_{i-1}+x_{i} \geqslant T_{i-2}+\sqrt{\frac{2}{3}}(\sqrt{i-1}+\sqrt{i})=T_{i}$$
Since either $n$ or $n+1$ is good by Lemma 2.1, Lemma 2.3 yields in both cases $S_{n+1} \geqslant T_{n}$, and it remains to apply Lemma 1.2 from Solution 1.