Let an index $i$ be good if $x_i \geq \sqrt{\frac{2i}{3}}$ and bad otherwise. Claim 1: For all $n$, $\sqrt{1}+\ldots+\sqrt{n} > \frac{2}{3}\sqrt{n^3}$. Proof: Induction works. For $n=1$ it is trivial and if it holds for $n$ then $\sqrt{1}+\ldots+\sqrt{n+1} \geq \frac{2}{3}\sqrt{n^3}+\sqrt{n+1}>\frac{2}{3}\sqrt{(n+1)^3},$ where the last inequality is a straightforward check $\blacksquare$ Claim 2: If $i$ is bad, then $i+1$ is good. Proof: Assume that $i+1$ is bad, too. Then, $x_{i-1}^2+1 \leq x_ix_{i+1} <\frac{2}{3}\sqrt{i(i+1)}<\frac{2}{3}\sqrt{(i+1/2)^2}=\frac{2}{3}(i+1/2)=\frac{2}{3}(i-1)+1,$ and so $i-1$ is bad, too. Continuing this way, we obtain that all $j \leq i$ are bad, and so we obtain a contradiction since $0$ is clearly good, as all $x_i$ are non-negative $\blacksquare$ Now, define $y_i=x_i-\sqrt{\frac{2i}{3}}$, and so $i$ is good if and only if $y_i \geq 0$. We prove the following Claim: Claim 3: If $i$ is good, then $y_{i+1}+y_{i+2} \geq 0$. Proof: Note that $x_{i+1}+x_{i+2} \geq 2\sqrt{x_{i+1}x_{i+2}} \geq 2\sqrt{x_i^2+1} \geq 2\sqrt{\frac{2i}{3}+1} \geq \sqrt{\frac{2(i+1)}{3}}+\sqrt{\frac{2(i+2)}{3}},$ where the last inequality follows from Jensen's inequality since $f(x)=\sqrt{\frac{2x}{3}+1}$ is concave. Thus, $y_{i+1}+y_{i+2} \geq 0$, as desired $\blacksquare$ Now, from Claim 1, we just need to prove that $x_0+\ldots+x_{n+1}>\sqrt{\frac{2}{3}}(\sqrt{1}+\ldots+\sqrt{n})$ This rearranges to $y_1+\ldots+y_n+x_{n+1}+x_0>0$ Claim 4: $y_1+\ldots+y_n+x_{n+1}+x_0 \geq \max \{y_1+\ldots+y_n,y_1+\ldots+y_{n+1} \}$ Proof: Indeed, $x_i \geq 0$ for all $i$ and $x_{n+1}+x_0 \geq x_{n+1} \geq y_{n+1}$. Combining these two, we are done $\blacksquare$ Since by Claim 2, for all $i$, either $i$ or $i+1$ is good, in order to finish we just need the following: Claim 5: If $i$ is good, $y_1+\ldots+y_i \geq 0$. This will finish if we use Claim 4. Proof: For $i=0$ it trivially holds, so let $i$ be a good index and assume that the inequality holds for all good indeces less than $i$. If $i-1$ is good then $y_1+\ldots+y_{i} \geq y_i \geq 0,$ as desired, while if $i-1$ is bad then $i-2$ must be good, by Claim 2, and so Claim 3 with the inductive hypothesis imply that $y_1+\ldots+y_i \geq y_{i-1}+y_i \geq 0,$ and we are done. $\blacksquare$

Claim: For any real numbers $x,y,z$ such that $xy-z^2\geq 1$, we have $(x+2y)^2\geq (y+2z)^2+6$. Proof. \begin{align*} (x+2y)^2\geq (y+2z)^2+6 &\iff (x^2+4xy+4y^2)-(y^2+4yz+4z^2)\geq 6 \\ &\iff (x^2-2xy+y^2)+(2y^2-4yz+2z^2)+(6xy-6z^2)\geq 6 \\ &\iff (x-y)^2+2(y-z)^2+6(xy-z^2)\geq 6 &\blacksquare \end{align*} Since $x_ix_{i+1}-x_{i-1}^2\geq 1$, we have $(x_{i+1}+2x_{i})^2\geq (x_{i}+2x_{i-1})^2+6$ for $i=1,2,\dots,n$. By induction, we have \begin{align*} x_1+2x_0 &\geq 0 \\ x_2+2x_1 &\geq \sqrt{6} \\ x_3+2x_2 &\geq \sqrt{12} \\ &\vdots \\ x_{n+1}+2x_{n} &\geq \sqrt{6n} \end{align*}Summing all of these equations give us that $$3\left(\sum_{i=0}^{n+1}x_i\right)\geq 2x_0+3x_1+3x_2+\cdots+3x_n+x_{n+1}\geq\sum_{i=1}^{n}\sqrt{6i}.$$Now, we're left to show that $\frac{\sum\limits_{i=1}^{n}\sqrt{6i}}{3}>\left(\frac{2n}{3}\right)^{\frac{3}{2}}$, which is true since $$\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}>\frac{2}{3}n^{\frac{3}{2}}.$$can be easily prove by integrating. This problem was proposed by Thailand.

Note that “equality” occurs at $x_i \approx \sqrt{2i/3}$. This motivates us to only utilize inequalities which compare nearby $x_i$’s, and to only use inequalities for which equality holds when the variables are equal. Now, sum all the inequalities and use $0.5x_i^2 + 0.5x_{i+1}^2\ge x_ix_{i+1}$ to obtain $x_1x_2 + x_2x_3+\dots + x_kx_{k+1} \ge k + x_0^2 + x_1^2 +\dots + x_{k-1}^2 \ge n + x_1x_2 + x_2x_3 + \dots + x_{k-2}x_{k-1} + 0.5x_{k-1}^2$. This implies $\implies x_{k-1}x_k + x_kx_{k+1} \ge k + 0.5x_{k-1}^2$. If we can use this result to prove a linear inequality in the $x_{k-1}, x_k, x_{k+1}$, then we can sum this inequality over all values of $k$ to finish. Since $x_{k-1}x_k \le 0.5x_{k-1}^2 + 0.5x_k^2$, we have $k \le x_kx_{k+1} + 0.5x_k^2$, so it follows that $1.5k \le 1.5x_k (x_{k+1} + 0.5x_k)$. Using $4ab\le (a+b)^2$ we thus obtain $6k \le (x_{k+1} + 0.5x_k + 1.5x_k)^2 = (x_{k+1} + 2x_k)^2$, hence $\sqrt{2k/3} \le \frac{1}{3} x_{k+1} + \frac{2}{3} x_k$. Summing over $k=1,2,\dots, n$, we thus obtain $x_0 + x_1 + \dots + x_{n+1} \ge \sqrt{2/3} (\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}) \ge \sqrt{2/3} \int_0^n \sqrt{x}\,dx = (2n/3)^{3/2}$ as desired.

Call an index $0 \le k \le n$ bad if $x_k < \sqrt{2k/3}$, and call it good otherwise. Note that $0$ is good. Consider any bad index $k \le n$ such that $k-1$ is good. Then \[x_{k+1} + x_k \ge \frac{x_{k-1}^2+1}{x_k} + x_k \ge \frac{2k+1}{3x_k} + x_k > \frac{2k+1}{3\sqrt{2k/3}} + \sqrt{2k/3} > \sqrt{2k/3} + \sqrt{2(k+1)/3}, \]where $\frac{2k+1}{3x_k} + x_k > \frac{2k+1}{3\sqrt{2k/3}} + \sqrt{2k/3}$ is true because $x_k < \sqrt{2k/3} < \sqrt{(2k+1)/3}$. This also means that $k+1$ is good. This actually shows that two consecutive indices cannot both be bad (otherwise let $j$ be the least index such that $j$ and $j+1$ are both bad. Then $j \ge 1$ and that $j-1$ must be good, but we just showed that if $j-1$ is good and $j$ is bad then $j+1$ is good). So, each bad index $1 \le k \le n$ can be paired with the good index $k+1$, and the sum $x_k + x_{k+1}$ is greater than $\sqrt{2k/3} + \sqrt{2(k+1)/3}$. The only unpaired bad index potentially is $n+1$. So, \[x_0 + x_1 + \ldots + x_{n+1} \ge \sum_{k=1}^{n} \sqrt{2k/3} > (2n/3)^{3/2}\]

Define $y_i=2x_i+x_{i+1}.$ We provide several inequalities as lemmas. Lemma 1. $\sum_{k=1}^n \sqrt{k} > \frac 23 n^{3/2}.$ Proof. Induct on $n;$ base case $n=1$ is trivial. The inductive step is provided by the Bernoulli's inequality $$(n+1)^{3/2}-n^{3/2}<(n+1)^{3/2}-(n+1)^{3/2}(1-\frac{3}{2(n+1)})<\frac 32 \sqrt {n+1}\text{ } \Box$$ Lemma 2. $y_i^2\geq y_{i-1}^2 +6.$ Proof. Rewrite as $y_i^2-y_{i-1}^2=6(x_{i+1}x_i-x_{i-1}^2)+(x_i-x_{i+1})^2+2(x_i-x_{i-1})^2\geq 6$ $\Box$ Lemma 3. $y_k\geq \sqrt {6k}.$ Proof. Induct on $k;$ for $k=0$ we have $y_0\geq 0.$ By hypothesis and Lemma 2 we have $y_{k+1}\geq \sqrt {6k+6}$ $\Box$ Now Lemmas 1,3 yield $\sum_{k=0}^{n+1} x_k \geq \frac 13 \sum_{k=1}^n y_k\geq \frac{\sqrt{6}}{3} \sum_{k=1}^n \sqrt {k}>\bigg(\frac{2n}{3}\bigg)^{3/2}$ $\blacksquare$

Let $f(i)=2x_i+x_{i+1}$ for $i\in \mathbb N$. We have: $$f(i)^2-f(i-1)^2=(x_{i+1}-x_i)^2+2(x_i-x_{i-1})^2+6(x_ix_{i+1}-x^2_{i-1})$$and so $f(i)^2\geqslant f(i-1)^2+6$, and induction here we get $f(i)\geqslant \sqrt{6i}$. Therefore, $3\sum_{i=0}^{n+1} x_i\geqslant \sum_{i=1}^n f(i)\geqslant \sqrt 6 \sum_{i=1}^n \sqrt i.$ But also $$\frac 32\sqrt a >\int_{a-1}^a \frac 32\sqrt x\; \text dx=\sqrt{a^3}-\sqrt{(a-1)^3}$$inducting on this we're done.

Working on it.

Note that $(x_{i+i}-x_i)^2+2(x_{i}-x_{i-1})^2+6(x_ix_{i+1}-x_i^2)\ge 6$, and this rearranges to $(x_{i+1}+2x_i)^2-(x_i+2x_{i-1})^2\ge 6$. This implies that $(x_{i+1}+2x_i)^2\ge 6i$, so \begin{align*} 3x_{n+1}+3x_{n-1}+\dots+3x_0 &> (x_{n+1}+2x_n)+(x_{n}+2x_{n-1})+\dots+(x_1+2x_0)\\ &\ge \sqrt{6n}+\sqrt{6(n-1)}+\dots+\sqrt{6} \end{align*}Which implies \[x_0+x_1+\cdots+x_n+x_{n+1}>\sqrt{\frac{2}{3}} \left(\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}\right)\]Note that $\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}$ is the Right Riemann Sum of $\sqrt{n}$ from $0$ to $n$ with increments of $1$, so \[\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}>\int_{0}^{n}{sqrt{n}}=\frac{2}{3}n^{\frac{3}{2}}\]and we are done.

Woah, nice problem! $(x_{i+1}-x_i)^2+2(x_i-x_{i-1})^2 + 6(x_ix_{i+1}-x_i^2) \geq 6$ so $(x_{i+1}+2x_{i})^2-(x_i+2x_{i-1})^2 \geq 6$ and therefore by trivial induction we have $x_i + 2x_{i-1} \geq \sqrt{6}\sqrt{i}$ so $\sum x_i \geq \sum \sqrt{n} \geq \bigg(\frac{2n}{3}\bigg)^{3/2}. $

I think this problem is actually very hard. Nice sollution.