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This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention.

For ease of reading, $$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[q]}=\sum_{i=1,i\ne p}^nx_i^q. \sum\text{ refers to }\sum_{i=1}^n.$$ Note that $S^q$ is not $S^{[q]}$ and $S_p^q$ is not $S_p^{[q]}$.

Define an equation $A_n$: $$\sum S_i^{[x_i]}=S^S.$$ For example, $A_3$ is: $${x_1}^{x_2}+{x_1}^{x_3}+{x_2}^{x_1}+{x_2}^{x_3}+{x_3}^{x_1}+{x_3}^{x_2}=({x_1}+{x_2}+{x_3})^{({x_1}+{x_2}+{x_3})}.$$

Without loss of generality, for every non-negative integer solutions (hereinafter called "solutions") for $A_n$, $x_i\le x_{i+1}$ for every $1\leq i<n$, then there are two distinct non-negetive solutions for $A_3$, one is ${x_1}=0,{x_2}=0,{x_3}=2$, and the other is ${x_1}=0,{x_2}=1,{x_3}=1$.

We call a solution for $A_n$ 'non-trivial' if $x_{n-1}\ne0$. The only known non-trivial solution is ${x_1}=0,{x_2}=1,{x_3}=1$ for $A_3$. The problem is: are there any more non-trivial solutions for $A_n$?
If so, please give an example.

Since this question is difficult enough, I will also recieve answers which give some features about every non-trivial solutions.

Update on 2021-06-26: Claim. For every solution to $A_n$, $$S^{S_n}\le n(n-1).$$

Proof.

1. Lemma 1. Claim. $$S_i^{[x_i]}\le(n-1)S_i^{x_i}.$$
   Proof. If $x_1=0$, then $$S_i^{[x_i]}=(n-1)S_i^{x_i}.$$ If $x_1\ne0$, then $$S_i^{[x_i]}\le S_i^{x_i}\le(n-1)S_i^{x_i}.$$
2. For every $1\le i<n$, $x_i\le x_{i+1}$, therefore, for every $1\le i\le n$, $x_i\le x_n$. And for every $1\le i\le n$, $x_i$ is non-negetive integer, therefore, for every $1\le i\le n$, $$S_i^{x_i}\le S_i^{x_n}\le S^{x_n}.$$
3. Therefore, $$S^S=\sum S_i^{[x_i]}\le(n-1)\sum S_i^{x_i}\le(n-1)\sum S^{x_n}=n(n-1)S^{x_n}, $$ that is, $$\frac{S^S}{S^{x_n}}\le n(n-1).$$ Since $$\frac{S^S}{S^{x_n}}=S^{S-x_n}=S^{S_n},$$ $$S^{S_n}\le n(n-1).$$ And that's what we want. $\tiny{\text{I've typed for an hour and I finished it finally :)}}$

Update on 2021-07-02: Claim. For every non-trivial solution to $A_n$, $$S^{S_n}\le\frac{n^2-3n+6}2.$$ Proof.

1. Split $x_i$ to $m$ zeros and $(n-m)$ non-zeros, $0\le m\le n-2$, $$S^S=\sum S_i^{[x_i]}=\sum_{i=1}^mS_i^{[x_i]}+\sum_{i=m+1}^nS_i^{[x_i]}\le m(n-1)+\sum_{i=m+1}^nS_i^{x_i}\le m(n-1)+(n-m)S^{x_n}.$$
2. $$S^{x_n}\ge n-m,$$ $$m(n-1)+(n-m)S^{x_n}\le\frac{m(n-1)}{n-m}\cdot S^{x_n}+(n-m)S^{x_n}=(n-m+\frac{m(n-1)}{n-m})S^x_n.$$ 3.$$d(n-m+\frac{m(n-1)}{n-m})/dx=\frac{n(n-1)}{(n-m)^2}-1,$$ which is always positive for $n-\sqrt{n(n-1)}\le m\le n-2$.
   We can see that $n-m+\frac{m(n-1)}{n-m}$ is the greatest when $m=n-2$, $$(n-m+\frac{m(n-1)}{n-m})S^x_n\le2+\frac{(n-2)(n-1)}2\cdot S^x_n=\frac{n^2-3n-6}2\cdot S^x_n.$$
3. $$S_S\le\frac{n^2-3n-6}2\cdot S^x_n,$$ $$S^{S_n}=\frac{S^S}{S^x_n}\le\frac{n^2-3n+6}2.$$ And that's what we want.

This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention.

For ease of reading, $$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[q]}=\sum_{i=1,i\ne p}^nx_i^q. \sum\text{ refers to }\sum_{i=1}^n.$$ Note that $S^q$ is not $S^{[q]}$ and $S_p^q$ is not $S_p^{[q]}$.

Define an equation $A_n$: $$\sum S_i^{[x_i]}=S^S.$$ For example, $A_3$ is: $${x_1}^{x_2}+{x_1}^{x_3}+{x_2}^{x_1}+{x_2}^{x_3}+{x_3}^{x_1}+{x_3}^{x_2}=({x_1}+{x_2}+{x_3})^{({x_1}+{x_2}+{x_3})}.$$

Without loss of generality, for every non-negative integer solutions (hereinafter called "solutions") for $A_n$, $x_i\le x_{i+1}$ for every $1\leq i<n$, then there are two distinct non-negetive solutions for $A_3$, one is ${x_1}=0,{x_2}=0,{x_3}=2$, and the other is ${x_1}=0,{x_2}=1,{x_3}=1$.

We call a solution for $A_n$ 'non-trivial' if $x_{n-1}\ne0$. The only known non-trivial solution is ${x_1}=0,{x_2}=1,{x_3}=1$ for $A_3$. The problem is: are there any more non-trivial solutions for $A_n$?
If so, please give an example.

Since this question is difficult enough, I will also recieve answers which give some features about every non-trivial solutions.

Update on 2021-06-26: Claim. For every solution to $A_n$, $$S^{S_n}\le n(n-1).$$

Proof.

1. Lemma 1. Claim. $$S_i^{[x_i]}\le(n-1)S_i^{x_i}.$$
   Proof. If $x_1=0$, then $$S_i^{[x_i]}=(n-1)S_i^{x_i}.$$ If $x_1\ne0$, then $$S_i^{[x_i]}\le S_i^{x_i}\le(n-1)S_i^{x_i}.$$
2. For every $1\le i<n$, $x_i\le x_{i+1}$, therefore, for every $1\le i\le n$, $x_i\le x_n$. And for every $1\le i\le n$, $x_i$ is non-negetive integer, therefore, for every $1\le i\le n$, $$S_i^{x_i}\le S_i^{x_n}\le S^{x_n}.$$
3. Therefore, $$S^S=\sum S_i^{[x_i]}\le(n-1)\sum S_i^{x_i}\le(n-1)\sum S^{x_n}=n(n-1)S^{x_n}, $$ that is, $$\frac{S^S}{S^{x_n}}\le n(n-1).$$ Since $$\frac{S^S}{S^{x_n}}=S^{S-x_n}=S^{S_n},$$ $$S^{S_n}\le n(n-1).$$ And that's what we want. $\tiny{\text{I've typed for an hour and I finished it finally :)}}$

Update on 2021-07-02: Claim. For every non-trivial solution to $A_n$, $$S^{S_n}\le\frac{n^2-3n+6}2.$$ Proof.

1. Split $x_i$ to $m$ zeros and $(n-m)$ non-zeros, $0\le m\le n-2$, $$S^S=\sum S_i^{[x_i]}=\sum_{i=1}^mS_i^{[x_i]}+\sum_{i=m+1}^nS_i^{[x_i]}\le m(n-1)+\sum_{i=m+1}^nS_i^{x_i}\le m(n-1)+(n-m)S^{x_n}.$$
2. $$S^{x_n}\ge n-m,$$ $$m(n-1)+(n-m)S^{x_n}\le\frac{m(n-1)}{n-m}\cdot S^{x_n}+(n-m)S^{x_n}=(n-m+\frac{m(n-1)}{n-m})S^{x_n}.$$
3. $$d(n-m+\frac{m(n-1)}{n-m})/dx=\frac{n(n-1)}{(n-m)^2}-1,$$ which is always positive for $n-\sqrt{n(n-1)}\le m\le n-2$.
   We can see that $n-m+\frac{m(n-1)}{n-m}$ is the greatest when $m=n-2$, $$(n-m+\frac{m(n-1)}{n-m})S^{x_n}\le2+\frac{(n-2)(n-1)}2\cdot S^{x_n}=\frac{n^2-3n-6}2\cdot S^{x_n}.$$
4. $$S^S\le\frac{n^2-3n-6}2\cdot S^{x_n},$$ $$S^{S_n}=\frac{S^S}{S^{x_n}}\le\frac{n^2-3n+6}2.$$ And that's what we want.

This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention.

For ease of reading, $$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[q]}=\sum_{i=1,i\ne p}^nx_i^q. \sum\text{ refers to }\sum_{i=1}^n.$$ Note that $S^q$ is not $S^{[q]}$ and $S_p^q$ is not $S_p^{[q]}$.

Define an equation $A_n$: $$\sum S_i^{[x_i]}=S^S.$$ For example, $A_3$ is: $${x_1}^{x_2}+{x_1}^{x_3}+{x_2}^{x_1}+{x_2}^{x_3}+{x_3}^{x_1}+{x_3}^{x_2}=({x_1}+{x_2}+{x_3})^{({x_1}+{x_2}+{x_3})}.$$

Without loss of generality, for every non-negative integer solutions (hereinafter called "solutions") for $A_n$, $x_i\le x_{i+1}$ for every $1\leq i<n$, then there are two distinct non-negetive solutions for $A_3$, one is ${x_1}=0,{x_2}=0,{x_3}=2$, and the other is ${x_1}=0,{x_2}=1,{x_3}=1$.

We call a solution for $A_n$ 'non-trivial' if $x_{n-1}\ne0$. The only known non-trivial solution is ${x_1}=0,{x_2}=1,{x_3}=1$ for $A_3$. The problem is: are there any more non-trivial solutions for $A_n$?
If so, please give an example.

Since this question is difficult enough, I will also recieve answers which give some features about every non-trivial solutions.

Update on 2021-06-26: Claim. For every solution to $A_n$, $$S^{S_n}\le n(n-1).$$

Proof.

1. Lemma 1. Claim. $$S_i^{[x_i]}\le(n-1)S_i^{x_i}.$$
   Proof. If $x_1=0$, then $$S_i^{[x_i]}=(n-1)S_i^{x_i}.$$ If $x_1\ne0$, then $$S_i^{[x_i]}\le S_i^{x_i}\le(n-1)S_i^{x_i}.$$
2. For every $1\le i<n$, $x_i\le x_{i+1}$, therefore, for every $1\le i\le n$, $x_i\le x_n$. And for every $1\le i\le n$, $x_i$ is non-negetive integer, therefore, for every $1\le i\le n$, $$S_i^{x_i}\le S_i^{x_n}\le S^{x_n}.$$
3. Therefore, $$S^S=\sum S_i^{[x_i]}\le(n-1)\sum S_i^{x_i}\le(n-1)\sum S^{x_n}=n(n-1)S^{x_n}, $$ that is, $$\frac{S^S}{S^{x_n}}\le n(n-1).$$ Since $$\frac{S^S}{S^{x_n}}=S^{S-x_n}=S^{S_n},$$ $$S^{S_n}\le n(n-1).$$ And that's what we want. $\tiny{\text{I've typed for an hour and I finished it finally :)}}$

Update on 2021-07-02: Claim. For every non-trivial solution to $A_n$, $$S^{S_n}\le\frac{n^2-3n-6}2.$$ Proof.

1. Split $x_i$ to $m$ zeros and $(n-m)$ non-zeros, $0\le m\le n-2$, $$S^S=\sum S_i^{[x_i]}=\sum_{i=1}^mS_i^{[x_i]}+\sum_{i=m+1}^nS_i^{[x_i]}\le m(n-1)+\sum_{i=m+1}^nS_i^{x_i}\le m(n-1)+(n-m)S^{x_n}.$$
2. $$S^{x_n}\ge n-m,$$ $$m(n-1)+(n-m)S^{x_n}\le\frac{m(n-1)}{n-m}\cdot S^{x_n}+(n-m)S^{x_n}=(n-m+\frac{m(n-1)}{n-m})S^x_n.$$ 3.$$d(n-m+\frac{m(n-1)}{n-m})/dx=\frac{n(n-1)}{(n-m)^2}-1,$$ which is always positive for $n-\sqrt{n(n-1)}\le m\le n-2$.
   We can see that $n-m+\frac{m(n-1)}{n-m}$ is the greatest when $m=n-2$, $$(n-m+\frac{m(n-1)}{n-m})S^x_n\le2+\frac{(n-2)(n-1)}2\cdot S^x_n=\frac{n^2-3n-6}2\cdot S^x_n.$$
3. $$S_S\le\frac{n^2-3n-6}2\cdot S^x_n,$$ $$S^{S_n}=\frac{S^S}{S^x_n}\le\frac{n^2-3n-6}2.$$ And that's what we want. $\tiny{\text{Aha! I did a bit better than @mathlove.}}$

This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention.

For ease of reading, $$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[q]}=\sum_{i=1,i\ne p}^nx_i^q. \sum\text{ refers to }\sum_{i=1}^n.$$ Note that $S^q$ is not $S^{[q]}$ and $S_p^q$ is not $S_p^{[q]}$.

Define an equation $A_n$: $$\sum S_i^{[x_i]}=S^S.$$ For example, $A_3$ is: $${x_1}^{x_2}+{x_1}^{x_3}+{x_2}^{x_1}+{x_2}^{x_3}+{x_3}^{x_1}+{x_3}^{x_2}=({x_1}+{x_2}+{x_3})^{({x_1}+{x_2}+{x_3})}.$$

Without loss of generality, for every non-negative integer solutions (hereinafter called "solutions") for $A_n$, $x_i\le x_{i+1}$ for every $1\leq i<n$, then there are two distinct non-negetive solutions for $A_3$, one is ${x_1}=0,{x_2}=0,{x_3}=2$, and the other is ${x_1}=0,{x_2}=1,{x_3}=1$.

We call a solution for $A_n$ 'non-trivial' if $x_{n-1}\ne0$. The only known non-trivial solution is ${x_1}=0,{x_2}=1,{x_3}=1$ for $A_3$. The problem is: are there any more non-trivial solutions for $A_n$?
If so, please give an example.

Since this question is difficult enough, I will also recieve answers which give some features about every non-trivial solutions.

Update on 2021-06-26: Claim. For every solution to $A_n$, $$S^{S_n}\le n(n-1).$$

Proof.

1. Lemma 1. Claim. $$S_i^{[x_i]}\le(n-1)S_i^{x_i}.$$
   Proof. If $x_1=0$, then $$S_i^{[x_i]}=(n-1)S_i^{x_i}.$$ If $x_1\ne0$, then $$S_i^{[x_i]}\le S_i^{x_i}\le(n-1)S_i^{x_i}.$$
2. For every $1\le i<n$, $x_i\le x_{i+1}$, therefore, for every $1\le i\le n$, $x_i\le x_n$. And for every $1\le i\le n$, $x_i$ is non-negetive integer, therefore, for every $1\le i\le n$, $$S_i^{x_i}\le S_i^{x_n}\le S^{x_n}.$$
3. Therefore, $$S^S=\sum S_i^{[x_i]}\le(n-1)\sum S_i^{x_i}\le(n-1)\sum S^{x_n}=n(n-1)S^{x_n}, $$ that is, $$\frac{S^S}{S^{x_n}}\le n(n-1).$$ Since $$\frac{S^S}{S^{x_n}}=S^{S-x_n}=S^{S_n},$$ $$S^{S_n}\le n(n-1).$$ And that's what we want. $\tiny{\text{I've typed for an hour and I finished it finally :)}}$

Update on 2021-07-02: Claim. For every non-trivial solution to $A_n$, $$S^{S_n}\le\frac{n^2-3n+6}2.$$ Proof.

1. Split $x_i$ to $m$ zeros and $(n-m)$ non-zeros, $0\le m\le n-2$, $$S^S=\sum S_i^{[x_i]}=\sum_{i=1}^mS_i^{[x_i]}+\sum_{i=m+1}^nS_i^{[x_i]}\le m(n-1)+\sum_{i=m+1}^nS_i^{x_i}\le m(n-1)+(n-m)S^{x_n}.$$
2. $$S^{x_n}\ge n-m,$$ $$m(n-1)+(n-m)S^{x_n}\le\frac{m(n-1)}{n-m}\cdot S^{x_n}+(n-m)S^{x_n}=(n-m+\frac{m(n-1)}{n-m})S^x_n.$$ 3.$$d(n-m+\frac{m(n-1)}{n-m})/dx=\frac{n(n-1)}{(n-m)^2}-1,$$ which is always positive for $n-\sqrt{n(n-1)}\le m\le n-2$.
   We can see that $n-m+\frac{m(n-1)}{n-m}$ is the greatest when $m=n-2$, $$(n-m+\frac{m(n-1)}{n-m})S^x_n\le2+\frac{(n-2)(n-1)}2\cdot S^x_n=\frac{n^2-3n-6}2\cdot S^x_n.$$
3. $$S_S\le\frac{n^2-3n-6}2\cdot S^x_n,$$ $$S^{S_n}=\frac{S^S}{S^x_n}\le\frac{n^2-3n+6}2.$$ And that's what we want.

This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention.

For ease of reading, $$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[q]}=\sum_{i=1,i\ne p}^nx_i^q. \sum\text{ refers to }\sum_{i=1}^n.$$ Note that $S^q$ is not $S^{[q]}$ and $S_p^q$ is not $S_p^{[q]}$.

Define an equation $A_n$: $$\sum S_i^{[x_i]}=S^S.$$ For example, $A_3$ is: $${x_1}^{x_2}+{x_1}^{x_3}+{x_2}^{x_1}+{x_2}^{x_3}+{x_3}^{x_1}+{x_3}^{x_2}=({x_1}+{x_2}+{x_3})^{({x_1}+{x_2}+{x_3})}.$$

Without loss of generality, for every non-negative integer solutions (hereinafter called "solutions") for $A_n$, $x_i\le x_{i+1}$ for every $1\leq i<n$, then there are two distinct non-negetive solutions for $A_3$, one is ${x_1}=0,{x_2}=0,{x_3}=2$, and the other is ${x_1}=0,{x_2}=1,{x_3}=1$.

We call a solution for $A_n$ 'non-trivial' if $x_{n-1}\ne0$. The only known non-trivial solution is ${x_1}=0,{x_2}=1,{x_3}=1$ for $A_3$. The problem is: are there any more non-trivial solutions for $A_n$?
If so, please give an example.

Since this question is difficult enough, I will also recieve answers which give some features about every non-trivial solutions.

Update on 2021-06-26: Claim. For every solutions to $A_n$, $$S^{S_n}\le n(n-1).$$

Proof.

1. Lemma 1. Claim. $$S_i^{[x_i]}\le(n-1)S_i^{x_i}.$$
   Proof. If $x_1=0$, then $$S_i^{[x_i]}=(n-1)S_i^{x_i}.$$ If $x_1\ne0$, then $$S_i^{[x_i]}\le S_i^{x_i}\le(n-1)S_i^{x_i}.$$
2. For every $1\le i<n$, $x_i\le x_{i+1}$, therefore, for every $1\le i\le n$, $x_i\le x_n$. And for every $1\le i\le n$, $x_i$ is non-negetive integer, therefore, for every $1\le i\le n$, $$S_i^{x_i}\le S_i^{x_n}\le S^{x_n}.$$
3. Therefore, $$S^S=\sum S_i^{[x_i]}\le(n-1)\sum S_i^{x_i}\le(n-1)\sum S^{x_n}=n(n-1)S^{x_n}, $$ that is, $$\frac{S^S}{S^{x_n}}\le n(n-1).$$ Since $$\frac{S^S}{S^{x_n}}=S^{S-x_n}=S^{S_n},$$ $$S^{S_n}\le n(n-1).$$ And that's what we want. $\tiny{\text{I've typed for an hour and I finished it finally :)}}$

This problem was first asked at Mathematics Stack Exchange, where it wasn't drawn much attention.

For ease of reading, $$S=\sum_{i=1}^nx_i, S_p=\sum_{i=1,i\ne p}^nx_i, S^{[q]}=\sum_{i=1}^nx_i^q, S_p^{[q]}=\sum_{i=1,i\ne p}^nx_i^q. \sum\text{ refers to }\sum_{i=1}^n.$$ Note that $S^q$ is not $S^{[q]}$ and $S_p^q$ is not $S_p^{[q]}$.

Define an equation $A_n$: $$\sum S_i^{[x_i]}=S^S.$$ For example, $A_3$ is: $${x_1}^{x_2}+{x_1}^{x_3}+{x_2}^{x_1}+{x_2}^{x_3}+{x_3}^{x_1}+{x_3}^{x_2}=({x_1}+{x_2}+{x_3})^{({x_1}+{x_2}+{x_3})}.$$

Without loss of generality, for every non-negative integer solutions (hereinafter called "solutions") for $A_n$, $x_i\le x_{i+1}$ for every $1\leq i<n$, then there are two distinct non-negetive solutions for $A_3$, one is ${x_1}=0,{x_2}=0,{x_3}=2$, and the other is ${x_1}=0,{x_2}=1,{x_3}=1$.

We call a solution for $A_n$ 'non-trivial' if $x_{n-1}\ne0$. The only known non-trivial solution is ${x_1}=0,{x_2}=1,{x_3}=1$ for $A_3$. The problem is: are there any more non-trivial solutions for $A_n$?
If so, please give an example.

Since this question is difficult enough, I will also recieve answers which give some features about every non-trivial solutions.

Update on 2021-06-26: Claim. For every solution to $A_n$, $$S^{S_n}\le n(n-1).$$

Proof.

1. Lemma 1. Claim. $$S_i^{[x_i]}\le(n-1)S_i^{x_i}.$$
   Proof. If $x_1=0$, then $$S_i^{[x_i]}=(n-1)S_i^{x_i}.$$ If $x_1\ne0$, then $$S_i^{[x_i]}\le S_i^{x_i}\le(n-1)S_i^{x_i}.$$
2. For every $1\le i<n$, $x_i\le x_{i+1}$, therefore, for every $1\le i\le n$, $x_i\le x_n$. And for every $1\le i\le n$, $x_i$ is non-negetive integer, therefore, for every $1\le i\le n$, $$S_i^{x_i}\le S_i^{x_n}\le S^{x_n}.$$
3. Therefore, $$S^S=\sum S_i^{[x_i]}\le(n-1)\sum S_i^{x_i}\le(n-1)\sum S^{x_n}=n(n-1)S^{x_n}, $$ that is, $$\frac{S^S}{S^{x_n}}\le n(n-1).$$ Since $$\frac{S^S}{S^{x_n}}=S^{S-x_n}=S^{S_n},$$ $$S^{S_n}\le n(n-1).$$ And that's what we want. $\tiny{\text{I've typed for an hour and I finished it finally :)}}$

Update on 2021-07-02: Claim. For every non-trivial solution to $A_n$, $$S^{S_n}\le\frac{n^2-3n-6}2.$$ Proof.

1. Split $x_i$ to $m$ zeros and $(n-m)$ non-zeros, $0\le m\le n-2$, $$S^S=\sum S_i^{[x_i]}=\sum_{i=1}^mS_i^{[x_i]}+\sum_{i=m+1}^nS_i^{[x_i]}\le m(n-1)+\sum_{i=m+1}^nS_i^{x_i}\le m(n-1)+(n-m)S^{x_n}.$$
2. $$S^{x_n}\ge n-m,$$ $$m(n-1)+(n-m)S^{x_n}\le\frac{m(n-1)}{n-m}\cdot S^{x_n}+(n-m)S^{x_n}=(n-m+\frac{m(n-1)}{n-m})S^x_n.$$ 3.$$d(n-m+\frac{m(n-1)}{n-m})/dx=\frac{n(n-1)}{(n-m)^2}-1,$$ which is always positive for $n-\sqrt{n(n-1)}\le m\le n-2$.
   We can see that $n-m+\frac{m(n-1)}{n-m}$ is the greatest when $m=n-2$, $$(n-m+\frac{m(n-1)}{n-m})S^x_n\le2+\frac{(n-2)(n-1)}2\cdot S^x_n=\frac{n^2-3n-6}2\cdot S^x_n.$$
3. $$S_S\le\frac{n^2-3n-6}2\cdot S^x_n,$$ $$S^{S_n}=\frac{S^S}{S^x_n}\le\frac{n^2-3n-6}2.$$ And that's what we want. $\tiny{\text{Aha! I did a bit better than @mathlove.}}$