By rearrangement, we need $$\sum_{i = 1}^{2020} ((x_i + 1)(x_{2021 - i} + 1))^2 \le 8 \sum_{i = 1}^{2020} x_i^3.$$ Consider the inequality $$(a + 1)^2(b + 1)^2 \le 4(a^3 + b^3),$$with $a \le b$. It can be shown that it is only true when $(a, b) = (0, 1), (1, 2)$; we do it using calculus. Evidently $b \in [a, a + 1]$, so $(a + 1)^2(b + 1)^2 - 4(a^3 + b^3)$, taking as a function of $b$, is minimized either when the derivative $$\frac{\partial}{\partial b} (a + 1)(b + 1)^2 - 4(a^3 + b^3) = 2(b + 1)(a + 1)^2 - 12b^2 = 0$$or at the endpoints of the interval. The derivative is a quadratic with roots at $$b = \frac{(a + 1)^2 \pm \sqrt{(a + 1)^4 + 24(a + 1)^2}}{12},$$and since $b \ge 0$, we must take the larger root. However, since we need a minimum, the second derivative at this root, $$\frac{\partial}{\partial b} 2(b + 1)(a + 1)^2 - 12b^2 = 2(a + 1)^2 - 24b,$$must be positive and thus $b < \frac{(a + 1)^2}{12}$. However $\frac{(a + 1)^2 + \sqrt{(a + 1)^4 + 24(a + 1)^2}}{12} > \frac{(a + 1)^2}{12}$, so there are no minimums inside the interval. If $b = a$, then we get $$(a + 1)^2 (b + 1)^2 - 4(a^3 + b^3) = (a + 1)^4 - 8a^3 = (a - 1)^4 + 8a,$$which is always positive, contradiction. If $b = a + 1$, then we get $$(a + 1)^2(a + 2)^2 - 4(a^3 + (a + 1)^3) = a^2(a - 1)^2,$$so as desired the only pairs $(a, b)$ where this is true are $(0, 1)$ and $(1, 2)$, and furthermore they are equality cases. Thus we must have either $$x_1 = x_2 = \ldots = x_{1010} = 0, x_{1011} = x_{1012} = \ldots = x_{2020} = 1,$$or $$x_1 = x_2 = \ldots = x_{1010} = 1, x_{1011} = x_{1012} = \ldots = x_{2020} = 2.$$

alifenix- wrote: Consider the inequality $$(a + 1)^2(b + 1)^2 \le 4(a^3 + b^3),$$with $a \le b$. It can be shown that it is only true when $(a, b) = (0, 1), (1, 2)$; we do it using calculus. Seems the following is cleaner (remembering that $|a-b| \le 1$ is given): \[ a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)[(a-b)^2 + ab] \]\[ \le (a+b)(1+ab) \le \left( \frac{(a+b)+(1+ab)}{2} \right)^2 = \frac{(a+1)^2(b+1)^2}{4}. \]

v_Enhance wrote: alifenix- wrote: Consider the inequality $$(a + 1)^2(b + 1)^2 \le 4(a^3 + b^3),$$with $a \le b$. It can be shown that it is only true when $(a, b) = (0, 1), (1, 2)$; we do it using calculus. Seems the following is cleaner: \[ a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)[(a-b)^2 + ab] \]\[ \le (a+b)(1+ab) \le \left( \frac{(a+b)+(1+ab)}{2} \right)^2 = \frac{(a+1)^2(b+1)^2}{4}. \] Oh yeah. Thanks. I forgot about that!

Isn't it given that $|a-b| \le 1$ for any $a$ and $b$ in the problem statement? (I suppose the quote I had didn't mention that )

Personally, i think this is a lot easier than 01, solved #01 much slower than this? Since there exists a permutation $\{y_i\}$ that satisfies such condition. By rearrangement Inequality, we have \[ \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021 - i} + 1))^2 \le 8 \sum_{i = 1}^{2020} x_i^3 \]$\textbf{Main Claim.}$ For any nonnegative reals $x \ge y$ such that $x - y \le 1$, then \[ 4(x^3 + y^3) \le ((x+1)(y+1))^2 \]$\textit{Proof.}$ Notice that \[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x+y)((x-y)^2 + xy) \le (x+y)(xy + 1) \le \frac{((x+1)(y+1))^2}{4} \]Equality holds if and only if $xy + 1 = x + y$ and $x = y + 1$, which gives us $(2,1)$ and $(1,0)$. Therefore, applying this to the original statement, we have that $x_i \{ 0, 1, 2 \}$ for each $i \in [1,2020]$. Now, suppose that $x_1 = 0$, this gives us $x_{2020} = 1$, forcing all the others $(x_i, x_{2020 - i})$ to be $(0,1)$ as well, giving $(0,0,0,\dots,0,1,1,\dots,1)$ as a solution. Suppose that $x_1 = 1$, this gives $x_{2020} = 2$, forcing all the others $(x_i, x_{2020 - i})$ to be $(1,2)$ as well, giving $(1,1,1,\dots,1,2,2,\dots,2)$ as a solution. Therefore, the solutions are \[ x_1 = x_2 = \dots = x_{1010} = 1 \ \text{and} \ x_{1011} = x_{1012} = \dots = x_{2020} = 2 \]and \[ x_1 = x_2 = \dots = x_{1010} = 0 \ \text{and} \ x_{1011} = x_{1012} = \dots = x_{2020} = 1 \] Remark. At first i tried to use the fact $|x - y| \le 1$ to prove $(x-1)^4 + (y - 1)^4 + 8(x+y) \ge (x+y+2)^2$, but when i square the initial condition (lol) and look back at the identity $x^3 + y^3$, solved this really fast accidentally.

Yeah. I didn't solve this very fast. Took me wayyyyy too much time (like > 30 minutes): We have the following Claim. For $0\leq x\leq y\leq x+1$, we have $4(x^3+y^3)\leq (x+1)^2(y+1)^2$. Furthermore, equality holds for $(x,y)=(0,1),(1,2)$. Proof. We start with defining the function \[f(y)=(x+1)^2(y+1)^2-4x^3-4y^3=-4y^3+(x+1)^2(y^2+2y+1)-4x^3\]Now, we expand to get \[f(y)=-4y^3+(x+1)^2y^2+2(x+1)^2y+(x+1)^2-4x^3\]Now, we note that \[f'(y)=-12y^2+2(x+1)^2y+2(x+1)^2\]We note that \[f'(x)=-12x^2+2(x+1)^2x+2(x+1)^2=2x^3-6x^2+6x+2=2(x-1)^3+4\]and \[f'(x+1)=-12(x+1)^2+2(x+1)^3+2(x+1)^2=2(x+1)^2(x-4)=2x^3-4x^2-14x-8\]Now, if $x\geq 4$, then $f(x)$ is decreasing, so it suffices to check $f(x)$ on its bounds. For $x<4$, we get that there is exactly one root in $[x,x+1]$ of $f'(x)$. We can check this is a maximum as \[f'\left(\dfrac{(x+1)^2}{12}\right)=-12\left(\dfrac{(x+1)^2}{12}\right)^2+2(x+1)^2\left(\dfrac{(x+1)^2}{12}\right)+2(x+1)^2\]Simplifying the first two, we get \[f'\left(\dfrac{(x+1)^2}{12}\right)=\dfrac{(x+1)^4}{12}+2(x+1)^2>0\]so thus we get that the root of $f'(x)$ is greater than $\dfrac{(x+1)^2}{12}$. Now, we get that if this root is $r$, it is a maximum if and only if $f''(r)>0$. However, we have \[f''(r)=-24r+(x+1)^2>0\]as $r>\dfrac{(x+1)^2}{12}$. Thus, in both cases where $x\geq 4$ and $x<4$, we have that the minimum of $f(x)$ occur on the bounds. Now, the rest is easy to check. We get \[f(x)=(x+1)^4-8x^3=(x-1)^4+8x>0\]as we can easily check that $(x+1)^4-(x-1)^4=8x^3+8x$. Thus, we also have \[f(x+1)=(x+1)^2(x+2)^2-4x^3-4(x+1)^3=(x+1)^2(x^2)-4x^3=(x-1)^2x^2\geq 0\]with equality only for $x=0,1$. Thus, the claim has been proved. $\blacksquare$. Now, we note by the Rearrangement Inequality, we get \[4\sum_{i=1}^{2020}(x_i^3+x_{2021-i}^3)=8\sum_{i=1}^{2020}x_i^3=\sum_{i=1}^{2020}(x_i+1)^2(y_i+1)^2\geq\sum_{i=1}^{2020}(x_i+1)^2(x_{2021-i}+1)^2\]However, we have that applying our claim on $x_i,x_{2020-i}$ for all $1\leq i\leq 2020$, we get \[4\sum_{i=1}^{2020}(x_i^3+x_{2021-i}^3\leq \sum_{i=1}^{2020}(x_i+1)^2(x_{2021-i}+1)^2\]with equality only for $x_1=x_2=\cdots=x_{1010}=0,1$ and $x_{1011}=x_{1012}=\cdots=x_{2020}$. Thus, we can define it as: \[\boxed{x_k=\begin{cases}0,1&k=1\\x_1&2\leq k\leq 1010\\x_1+1&1011\leq k\leq 2020\end{cases}}\]I hate not being able to see like v_enhance.

Here is a cleaner way to prove the inequality. The answer is $(0,0,\hdots,0,1,1,\hdots,1)$ and $(1,1,\hdots,1,2,2,\hdots,2)$ where each numbers appear $1010$ times. This is easily seen to work so we prove that these are all solutions. Consider the following. Claim: For any reals $a,b$ such that $|a-b|\leq 1$, we have $(a+1)^2(b+1)^2\geq 4(a^3+b^3)$, with equality at $(0,1)$, $(1,2)$. Proof: We first reduce this to one-var. Set $a=t+\delta$ and $b=t-\delta$ where $\delta\in (0,\tfrac 12]$. We have $$\text{LHS} = ((t+1)^2-\delta^2)^2 \geq \left((t+1)^2-\tfrac{1}{4}\right)^2.$$Moreover, $$\text{RHS} = 4((t+\delta)^3 + (t-\delta)^3)= 8t(t^2+3\delta^2) \leq 8t\left(t^2+\tfrac{3}{4}\right)$$Hence it suffices to prove that $\left((t+1)^2-\tfrac{1}{4}\right)^2\geq 8t^3+6t$, which could be easily expanded to $\left(t-\tfrac{1}{2}\right)^2\left(t-\tfrac{3}{2}\right)^2\geq 0$. The equality case can easily be reverse-engineered. $\blacksquare$ Applying the claim with $(x_i, y_i)$ and sum up, we get that $$\sum_{i=1}^{2020}((x_i+1)(y_i+1))^2\geq 8\sum_{i=1}^{2020}x_i^3.$$The equality case occurs if and only if $(x_i, y_i) = (0,1), (1,2)$ for each $i$. It's easy to see that they must be $(0,1)$ or $(1,2)$ for all $i$. Moreover, each number appears in $x_1,x_2,\hdots,x_{2020}$ exactly $1010$ times. Hence there are two such lists.

It's enough to prove that $(a+1)^2(b+1)^2\geq4(a^3+b^3),$ where $a$ and $b$ are non-negatives such that $(a-b)^2\le1.$ Indeed, let $a+b=2u$ and $ab=v^2$. Thus, $$4u^2-4v^2\leq1$$and we need to probe that $$(v^2+2u+1)^2\geq4(8u^3-6uv^2)$$or $$v^4+(28u+2)v^2+(2u+1)^2\geq32u^3,$$for which it's enough to prove that $$v^4+(28u+2)v^2+(2u+1)^2\geq8u(4v^2+1)$$or $$v^4-2(2u-1)v^2+(2u-1)^2\geq0$$or $$(v^2-2u+1)^2\geq0$$and from here we can get the equality cases occurring.

Solved with TheUltimate123 and eisirrational. The answer is $(\underbrace{0,\dots,0}_{1010 \ \text{0's}}, \underbrace{1,\dots,1}_{1010 \ \text{1's}})$ and $(\underbrace{1,\dots,1}_{1010 \ \text{1's}}, \underbrace{2,\dots,2}_{1010 \ \text{2's}})$. These can be easily checked to work, so we now show uniqueness. By the Rearrangement Inequality, we have that \[8 \sum_{i = 1}^{2020} x_i^3 = \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 \geq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2.\] We, in fact, show the reverse inequality in the following claim. Claim. For all reals $a \leq b \leq a+1$, we have $8(a^3+b^3) \leq 2(a+1)^2(b+1)^2$. Moreover, equality holds when $(a,b) = (0,1)$ or $(1,2)$. Proof. Verify that \[(a+1)^2(b+1)^2 - 4(a^3+b^3) = (a-1)^2(b-1)^2 + 4(a+b)(1-(a-b)^2) \geq 0\]as desired. Finally, apply the claim to $(a,b) = (x_i, x_{2021-i})$ and sum the resulting inequalities to show that \[8 \sum_{i = 1}^{2020} x_i^3 \leq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2.\]So equality must hold, as desired.

Claim: If $0\le b-a\le 1$, then $(a+1)^2(b+1)^2 \ge 4(a^3+b^3)$, with equality iff $(a,b)=(0,1),(1,2)$. Proof: We see that \begin{align*} 4(a^3+b^3) &= 4(a+b)(a^2-ab+b^2) = 4(a+b)[(b-a)^2 + ab] \\ &\le 4(a+b)(1+ab) = (2a+2b)(2+2ab) \le (a+b+1+ab)^2 \\ &=(a+1)^2(b+1)^2. \end{align*}Equality is achieved when $b-a=1$ and $a+b=1+ab$ simultaneously, i.e. $(a,b)=(0,1),(1,2)$. $\square$ By the Rearrangement inequality, the minimum LHS is achieved when we pair the smallest elements with the largest available: \[ \sum_{i=1}^{2020} ((x_i+1)(y_i+1))^2 \ge \sum_{i=1}^{2020} ((x_i+1)(x_{2021-i}+1))^2. \]Since all the $x_i$'s are within 1 of each other, we apply the claim to get \[ ((x_i+1)(x_{2021-i}+1))^2 \ge 4(x_i^3+x_{2021-i}^3). \]Sum the above cyclically to conclude that the LHS is at least the RHS in the third condition. Equality is achieved when $(x_i,x_{2021-i}) = (0,1)\text{ or }(1,2)$ for all $i$. Note that we cannot have both combinations in a single working tuple, since all the $x_i$'s are within 1 of each other. Hence, the two equality cases are \[ (x_1,\ldots,x_{2020}) = (0,\ldots,0,1,\ldots,1), \ (1,\ldots,1,2,\ldots,2), \]where each of the two tuples above has 1010 of each number.

The two lists satisfying these criteria are $(1,1,\dots, 1,2,2,\dots, 2)$ where there are $1010$ of each of $1$ and $2$ and the similarly defined $(0,0,\dots, 0,1,1,\dots, 1)$. It is easy to check that both of these work. Now, consider some satisfactory sequence. By Rearrangement Inequality, we have \[\sum_{i=1}^{2020} (x_i+1)^2(y_i+1)^2 \ge \sum_{i=1}^{2020} (x_i+1)^2(x_{2021-i}+1)^2 = 2\sum_{i=1}^{1010} (x_i+1)^2(x_{2021-i}+1)^2.\]Clearly, there exists some $i$ with \[(x_i+1)^2(x_{2021-i}+1)^2 \le 4(x_i^3+x_{2021-i}^3).\]Let $x_i+1=a,x_{2021-i}+1=b$. Note that $b \le 1+a$. We have \[a^2b^2\le 4(a^3-3a^2+3a-1+(b-1)^3).\]Observe that the function \[f(b) = a^2b^2-(b-1)^3\]has \[f’(b) = 2a^2b - 3(b-1)^2 \ge 2a^2b-3a^2 = a^2(2b-3),\]so $f$ certainly has nonnegative slope for $b \ge 3/2$. For $b \le 3/2$, we get $a^2b^2 \ge 9/4$ and $4(a-1)^3 + 4(b-1)^3 \le 1$, a contradiction. Thus, we ought to check this inequality at just $b=a+1$ and $b=a$. At $b=a$ we get \[a^4\le 8(a-1)^3 = (2a-2)^3.\]Note that the function $f(a) = a^4-(2a-2)^3$ is convex because $f''(a) = 12a^2 - 48(a-1) = 12(a^2-4a+4)$ is nonnegative. Thus, this inequality can never be satisfied because $1^4 > 8(1-1)^3$ and $f’(1) = 4\cdot 1^3-24(1-1)^2 > 0$. That is, we now only need to check $b=a+1$. Plug in $b=a+1$ to obtain \[a^4+2a^3+a^2 = a^2(a+1)^2 \le 4(a-1)^3 + 4a^3 = 8a^3 - 12a^2+12a-4.\]Rearrange to get \[0 \ge a^4-6a^3+13a^2-12a+4 = (a-2)(a^3-4a^2+5a-2) = (a-2)^2(a^2-2a+1) = (a-1)^2(a-2)^2.\]Note that this inequality is always at most an equality; this means that each $x_i+1$ is one of $1$ or $2$. But clearly the information about $b$ forces either all $x_i$ with $1\le i \le 1010$ to be $0$ or all $1$. This yields the two claimed solutions above.

My solution is same as everyone done . BUT MY Solution to $4(x^3+y^3)\le (x+1)^2(y+1)^2)$ is bit different . Here it is - $(x+1)^2(y+1)^2$ $=(xy+x+y+1)^2$ $\ge (xy+x+y +(x-y)^2)^2$ [cause $|x-y|\le 1$ ]. $=(x+y +(x^2+y^2-xy)^2$ $\ge (x+y)(x^2+y^2-xy).4$[AM-GM] $=4(x^3+y^3)$ . Obviously equality holds for $(x-y)^2=1$ and $x^2+y^2-xy =x+y$ .

So troll We claim that the only solutions are \[\boxed{x_k=\begin{cases}0,1&k=1\\x_1&2\leq k\leq 1010\\x_1+1&1011\leq k\leq 2020\end{cases}}\] Invoking Rearrangement inequality, we have \[8 \sum_{i = 1}^{2020} x_i^3 = \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 \geq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2.\]. We now present a crucial claim . $\mathbf{Lemma}$ Consider non-negative reals $a \leq b$ with $a-b \leq 1$ . We claim the following inequality holds $$(a + 1)^2(b + 1)^2 \geq 4(a^3 + b^3)$$ with equality iff $(a,b)=(1,0) \text { or} (2,1)$ $\mathbf{Proof :}$ We have $$\left((a+1)(b+1) \right)^2 = \left((a+b) +(ab+1)\right)^2 \geq 4(a+b)(ab+1) \geq 4(a+b)(ab+ (a-b)^2) = 4(a+b)(a^2+ab+b^2) = 4(a^3+b^3) $$Equality occurs when $a=b+1$ and $a+b=ab+1$ . Now we are ready to finish . Note that , from the claim we have, $$ \sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2 \geq 8\sum_{i = 1}^{2020} x_i^3 $$. So we have the forementioned solutions as desired as equality must hold in all the estimates we used . $\blacksquare$ .

We know that $(x_1+1)^2\leq (x_2+1)^2 \leq \cdots \leq (x_{2020}+1)^2 $ . Now Rearrangement inequality gives us $$\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 \geq \sum_{i = 1}^{2020} ((x_i + 1)(x_{2020-i} + 1))^2$$Now I'll prove that for each $1\leq i \leq 1010$ e have $$((x_i + 1)(x_{2020-i} + 1))^2+((x_{2020-i} + 1)(x_i + 1))^2 \geq 8(x_i^3+x_{2020-i}^3) (*)$$and when summing all of these up we would get $$\sum_{i = 1}^{2020} ((x_i + 1)(x_{2020-i} + 1))^2 \geq 8 \sum_{i = 1}^{2020} x_i^3 $$so the only solutions will be the equality cases in $(*)$. Now let's for brevity have $x_i=x$ and $x_{2020-i}=y$ and we also know that $x_{2020-i}-x_i=y-x\leq 1$. Now let $y-x=k$, so we know that $0\leq k\leq 1$. Now $(*)$ transforms into proving $$(x+1)^2(x+1+k)^2 \geq 4(x^3+(x+k)^3) \Leftrightarrow $$$$(x+1)^4+2(x+1)^3k+(x+1)^2k^2 \geq 4(x^3+x^3+3x^2k+3xk^2+k^3) \Leftrightarrow $$$$x^4+4x^3+6x^2+4x+1+2x^3k+6x^2k+6xk+2k+x^2k^2+2xk^2+k^2\geq $$$$\geq 8x^3+12x^2k+12xk^2=4k^3 \Leftrightarrow $$$$x^4+2x^3(k-2)+x^2(6-6k+k^2)+x(4+6k-10k^2)+(k^2+2k+1-4k^3)\geq 0$$$$x^2(x^2+2x(k-2)+(k-2)^2)+x^2(6-6k+k^2-(k-2)^2)+x(4+6k-10k^2)+(k^2+2k+1-4k^3)\geq 0$$$$x^2(x+(k-2))^2+2x^2(1-k)+x(k-1)(-10k-4)+(k-1)(-4k^2-3k-1)\geq 0$$Now we have the following inequalities (for $0\leq k\leq 1$) $$x^2(x+(k-2))^2\geq 0$$$$2x(1-k)\geq 0$$$$x(k-1)(-10k-4)\geq 0$$$$(k-1)(-4k^2-3k-1)\geq 0$$There is an equality in the last one only when $k=1$, so we get that we should have $k=1$.Now $(*)$ transforms into $$x^2(x-1)^2\geq 0$$and equality holds only when $x=0;1$. We got that in $$((x_i + 1)(x_{2020-i} + 1))^2+((x_{2020-i} + 1)(x_i + 1))^2 \geq 8(x_i^3+x_{2020-i}^3)$$equality holds only when $x_i=1$ and $x_{2020-i} = 2$ or when $x_i=0$ and $x_{2020-i}=1$. Together with the constraints in the statement we get that the only solutions are $$x_1=x_2=\cdots =x_{1010}=0 ; x_{1011}=\cdots = x_{2020}=1 $$$$x_1=x_2=\cdots =x_{1010}=1 ; x_{1011}=\cdots = x_{2020}=2 $$

Yet another proof of the main inequality The answer is $(0,\ldots,0,1,\ldots,1)$ and $(1,\ldots,1,2,\ldots,2)$, where both tuples contain exactly $1010$ ones. We can verify that these work by letting $(y_1,\ldots,y_{2020})$ be the reverse of $(x_1,\ldots,x_{2020})$. The key inequality is the following bound: $$\sum_{i=1}^{2020}(x_i+1)^2(x_{2020-i}+1)^2\geq 8\sum_{i=1}^{2020} x_i^3.$$It suffices to "break up" this summation and prove $$(x_i+1)^2(x_{2020-i}+1)^2\geq 4x_i^3+4x_{2020-i}^3.$$Let $a=x_i$ and $b=x_{2020-i}$, so $a \leq b \leq a+1$. The inequality becomes $(a+1)^2(b+1)^2 \geq 4a^3+4b^3 \iff (a+1)^2(b+1)^2-4a^3-4b^3 \geq 0$. Fix $b$ and let $f(a)$ denote the value of the expression that we want to prove is nonnegative over the interval $I=[0,1] \cap [b-1,b]$. We can compute $f'(a)=2(b+1)^2(a+1)-12a^2$. Since $f'(0)=2(b+1)^2>0$ and $f'(-\infty)=-\infty<0$, it follows that there is exactly one nonnegative root of $f'$ (Descartes' rule of signs works too!). Hence there is at most one turning point of $f$ in $I$, where $f$ must go from increasing to decreasing. As such, $f$ attains its minimum at an endpoint of $I$. We have \begin{align*} f(b)&=(b+1)^4-8b^3=b^4-4b^3+6b^2+4b+1=(b-1)^4+8b^3>0\\ f(b-1)&=b^2(b+1)^2-4b^3-4(b-1)^3=(b-2)^2(b-1)^2\geq 0\\ f(0)&=(b+1)^2-4b^3=(1-b)(4b^2+3b+1)\geq 0~\forall b \leq 1,\\ \end{align*}hence $f(a)$ is indeed nonnegative for $a \in I$, and is zero only if $(a,b)=(0,1)$ or $(a,b)=(1,2)$, by looking at the equality cases. Then by rearrangement, for any permutation $(y_1,\ldots,y_{2020})$, we have $$\sum_{i=1}^{2020}(x_i+1)^2(y_i+1)^2\geq \sum_{i=1}^{2020}(x_i+1)^2(x_{2020-i}+1)^2\geq 8\sum_{i=1}^{2020} x_i^3.$$It is clear that we can't have both of our equality cases hold at the same time (since that makes $x_1<x_{2020}+1$), so we extract the desired solution set. $\blacksquare$

The key is to use the following equaltiy: Claim. $$(x+1)^2(y+1)^2 \geq 4(x^3+y^3)$$as long as $|x-y| \leq 1$ with equality at $(0, 1)$ and $(1, 2)$. We can use the inequality $(x-y)^2 \leq 1$, and substitute $a = x+y, b = xy$, so that the given equation reduces to $$(a+b+1)^2 \geq 4a(b+1).$$This can just be done by considering $\Delta_a$ for the resulting quadratic in $a$, details omitted. $\blacksquare$ For the final part, we can use Rearrangement in the form $$\sum_{i=1}^{2020} (x_i+1)^2(y_i+1)^2 \geq \sum_{i=1}^{2020} (x_i+1)^2(x_{2021-i}+1)^2 \geq 8\sum_{i=1}^{1010} (x_i^3+x_{2021-i}^3).$$Thus, equality must hold everywhere, which implies that $(x_i, x_{2021-i}) = (0, 1)$ or $(1, 2)$. Obviously the solutions must all come from one of these combinations, so only $(0, 0, \cdots, 0, 1, 1, \cdots, 1)$ and $(1, 1, \cdots, 1, 2, 2, \cdots, 2)$, where there are 1010 of each number, work.

Our given condition can be rewritten as \[\sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 4 \sum_{i = 1}^{2020} (x_i^3 + y_i^3).\] Due to the order we are given, Rearrangement inequality implies \[\sum_{i = 1}^{2020} ((x_i + 1)(x_{2021-i} + 1))^2 \leq 4 \sum_{i = 1}^{2020} (x_i^3 + x_{2021-i}^3).\] However, comparing each individual summand, we notice the inequality \[\left((a+1)(b+1)\right)^2 \ge \left(2 \sqrt{(ab+1)(a+b)}\right)^2 = 4(a+b)(ab+1) \ge 4(a+b)(ab+(a-b)^2) = 4(a^3+b^3),\] where $a = x_i$ and $b = x_{2021-i}$ for simplicity, and $(a-b)^2 \leq 1$ by our second condition. Hence we must have the equality case, or \begin{align*} (a-b)^2 = 1 &\implies a = b \pm 1 \\ ab+1 &= a+b, \end{align*} from which we get the pairs $(0, 1)$ and $(1, 2)$. Thus our possible sequences are \[\boxed{a_1 = a_2 = \ldots = a_{1010} = 0, \quad a_{1011} = a_{1012} = \ldots = a_{2020} = 1}\]\[\boxed{a_1 = a_2 = \ldots = a_{1010} = 1, \quad a_{1011} = a_{1012} = \ldots = a_{2020} = 2}\]

The answer is $(\underbrace{0,\dots,0}_{1010 \ \text{0's}}, \underbrace{1,\dots,1}_{1010 \ \text{1's}})$ and $(\underbrace{1,\dots,1}_{1010 \ \text{1's}}, \underbrace{2,\dots,2}_{1010 \ \text{2's}})$. These can be easily checked to work, so now we prove that these are the only solutions Claim: For reals $0 \le x \le y \le x+1$, we have $2(x+1)^2(y+1)^2 \ge 8(x^3+y^3)$, equality at $(x,y)=(0,1), (1,2)$. Proof: Expansion gives \begin{align*} &(x+1)^2(y+1)^2-4(x^3+y^3) \\ = \ &(x-1)^2(y-1)^2+4(x+y)(xy+1)-4(x^3+y^3) \\ = \ &(x-1)^2(y-1)^2 + 4(x+y)(xy+1-(x^2+y^2-xy)) \\ = \ &(x-1)^2(y-1)^2 + 4(x+y)(1-(x-y)^2) \ge 0. \ \square \end{align*} Substituting in $(x,y)=(x_i, x_{2021-i})$ and summing, we get \[\sum_{i=1}^{2020} ((x_i-1)(x_{2021-i}-1))^2 \ge 8 \sum_{i = 1}^{2020} x_i^3\] However, by the Rearrangement Inequality, we have \[\sum_{i=1}^{2020} ((x_i-1)(x_{2021-i}-1))^2 \le \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3\] Thus, we must have the equality case, meaning $(x_i, x_{2021-i})=(0,1), (1,2)$ for $1 \le 1 \le 1010$, giving us our two solutions.

Note that \[\sum_{i=1}^{2020}{(x_i+1)^2(x_{2021-i}+1)^2} \leq \sum_{i=1}^{2020}{(x_i+1)^2(y_i+1)^2}\]from rearrangement inequality. We claim that \[8(x^3+y^3) \leq 2(x+1)^2(y+1)^2\]with equality case when they're both $1$, if $(x-y)^2\leq 1$. From, $(x-y)^2\leq 1$, we get \[(x-y)^2\leq 1 \implies x^2+y^2 \leq 1+2xy \implies x^3+y^3 \leq (1+xy)(x+y)\]Note that \[0\leq(x-1)^2(y-1)^2 \implies 4(1+xy)(x+y) \leq (x+1)^2(y+1)^2\]completing the proof. The equality case of such inequality is when one of them is $1$ and the other is $1$ away, giving $(0,1)$ and $(1,2)$. Thus, our only two sequences are $x_{i} = (0,1)$, $i \in \{1, 2, \cdots, 1010\}$, $x_{i} = (1,2)$, $i \in \{1011, 1012, \cdots, 2020\}$.

We first prove that if nonnegative $x,y$ satisfy $(x-y)^2\le 1,$ then $((x+1)(y+1))^2 \ge 4(x^3+y^3).$ To do this, let $x+y=s,xy=p$ and our condition is $s^2\le 4p+1.$ We wish to show that $s^2+p^2+14sp+2s+2p+1 \ge 4s^3.$ From our condition we may instead consider $s^2+p^2+14sp+2s+2p+1 \ge 16sp+4s,$ which implies the result. However this simplifies to $(p-s+1)^2 \ge 0,$ which is clear, and equality holds iff both $(x-y)^2=1$ and $(xy-x-y+1)^2=0,$ that is, either $x=1$ or $y=1.$ Thus equality holds when $(x,y)=(0,1),(1,0),(1,2),(2,1).$ Now by Rearrangement Inequality on our original expression, we see that $\sum_{i=1}^{2020}((x_i+1)(y_i+1))^2 \ge \sum_{i=1}^{2020}((x_i+1)(x_{2021-i}+1))^2.$ Now letting $x_i=x$ and $x_{2021-i}=y,$ since we have $|x_i-x_{2021-i}| \le |x_{2020}-x_1|\le 1,$ we see that $((x_i+1)(x_{2021-i}+1))^2 \ge 4(x_i^3+x_{2021-i}^3).$ Summing this across all $1 \le i \le 2020,$ we see that $\sum_{i=1}^{2020}((x_i+1)(x_{2021-i}+1))^2 \ge 8\sum_{i=1}^{2020} x_i^3,$ and equality holds iff each pair of $(x_i,x_{2021-i})$ is one of the four we found above. From here we notice that we cannot have both a $0$ and a $2,$ and since the sequence is increasing we get that the only possibilities are $x_1=x_2=\cdots=x_{1010}=0,x_{1011}=x_{1012}=\cdots=x_{2020}=1$ and $x_1=x_2=\cdots=x_{1010}=1,x_{1011}=x_{1012}=\cdots=x_{2020}=2.$

We will first show that $4(x^3 + y^3) \leq ((x + 1)(y + 1))^2$, if $|x-y| \leq 1$. \[(xy + x + y + 1)^2 = ((x + 1)(y + 1))^2\]\[= (xy + x + y + 1)^2 \geq (xy + x + y + (x - y)^2)^2 \]\[= (x^2 - xy + x + y + y^2)^2 \geq 4(x^2 - xy + y^2)(x + y) \]\[= 4(x^3 + y^3) \]\[\implies 4(x^3 + y^3) \leq ((x + 1)(y + 1))^2 \] Then by Rearrangement Inequality, $\newline$ \[\sum_{i = 1}^{2020} ((x_i + 1)(x_{2020-i} + 1))^2 \leq \sum_{i=1}^{2020} ((x_i + 1)(y_i + 1))^2 = \sum_{i = 1}^{2020} x_i^3 \]Which implies that this is the equality case of the Rearrangement Inequality, so the list $x_i$ either consists of $1010$ $0$'s and $1010$ $1$'s, or $1010$ $1$'s and $1010$ $2$'s.

Feels more like an equality problem instead of manipulation. This took way too long, why didn't I try the obvious thing to do first We will prove that the only solutions $(x_1, x_2, \dots, x_{2020})$ is the multiset containing $1010$ copies of $0$ and $1$, and the multiset containing $1010$ copies of $1$ and $2$, which are easily shown to work. We begin with the following claim: $\textbf{Claim 1.}$ If $|a - b| \le 1$, then $(a + 1)^2 (b + 1)^2 \le 4(a^3 + b^3)$. Proof. Let $a + b = 2u$, $ab = v^2$, we have \[|a - b| \le 1 \iff (a + b)^2 - 4ab \le 1 \iff v^2 \ge \dfrac{4u^2 - 1}{4}\]and we wish to show \[(a + 1)^2 (b + 1)^2 \ge 4(a^3 + b^3) \iff (ab + a + b + 1)^2 \ge 4(a + b)[(a + b)^2 - 3ab] \iff (2u + v^2 + 1)^2 \ge 8u(4u^2 - 3v^2). \]Now utilizing the earlier condition we have $(2u + v^2 + 1)^2 \ge \left( \dfrac{4u^2 + 8u + 3}{4} \right)^2$ and $8u \left( \dfrac{4u^2 + 3}{4}\right) \ge 8u(4u^2 - 3v^2)$ so it suffices to prove that \[\left( \dfrac{4u^2 + 8u + 3}{4} \right)^2 \ge 8u \left( \dfrac{4u^2 + 3}{4}\right) \iff 16u^4 - 64u^3 + 88u^2 - 48u + 9 \ge 0 \iff (2u - 1)^2 (2u - 3)^2 \ge 0\]which is true. Equality cases occur at $u = 1/2, 3/2$, and reverse-engineering the original pairs we get $(a, b) = (0, 1), (1, 2)$ and permutations. Now returning to the original problem, note that by Chebyshev (also extended Rearrangement) we obtain $$8 \sum_{i = 1}^{2020} x_i^3 \ge \sum_{i = 1}^{2020} (x_i + 1)^2 (x_{2021 - i} + 1)^2$$but summing $\textbf{Claim 1.}$ over all $i$ on ordered pairs $(x_i, x_{2021 - i})$ yields the same inequality with an inverted sign. Hence, equality must hold, which easily generates the solution sets claimed earlier.

What a missed solve ;-;

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