First solution, by me: Since the conditions are all homogeneous assume WLOG that $a_1=1$. FSoC assume that the sequence is not constant, so that we are able to let $a_1=a_2=\dots=a_k=1$ and $a_{k+1}\neq 1$. Conditions are immediately violated if $a_{k+1}>1$ so we must have $a_{k+1}<1$; write $a_{k+1}=1-d$ for some $d>0$. Squaring and rearranging the given inequality gives \[(n+1)\left(\frac{\sum_{i\leq n} a_i}n\right)^2-\sum_{i\leq n} a_i^2\geq a_{n+1}^2\]which becomes \[\left(\frac{\sum_{i\leq n} a_i}n\right)^2-\frac{\sum_{i<j\leq n}(a_i-a_j)^2}{n}\geq a_{n+1}^2\]by Lagrange's identity. Lemma 1: For all $n\geq k+1$, we have $a_n\leq 1-\frac d{k+1}$. Proof: The case $n=k+1$ is obvious as $a_{k+1}=1-d\leq 1-\frac d{k+1}$. Note that for all $n\geq k+2$, we inductively have \[a_n\leq \frac{\sum_{i\leq n-1}a_i}{n-1}\leq 1-\frac d{k+1}.\qquad\blacksquare\] Lemma 2: $\sqrt{\mu^2-x}\leq \mu -\frac x{2\mu}$. Proof: Squaring both sides yields \[\mu^2-x\leq \mu^2-x+\frac{x^2}{4\mu^2}\]which is obvious. $\blacksquare$ Now, we are ready for the key step. Taking \[\mu = \frac{\sum_{i\leq n} a_i}{n}\]and \[x=\frac{\sum_{i<j\leq n}(a_i-a_j)^2}{n}\]yields \[a_{n+1}\leq \sqrt{\mu^2-x}\leq \mu-\frac x{2\mu} = \frac{\sum_{i\leq n} a_i}{n}-\frac{\sum_{i<j\leq n}(a_i-a_j)^2}{2\sum_{i\leq n} a_i}.\] Therefore, as we increment $n$ for each $n\geq k+1$, the average of the first $n$ terms decreases by \begin{align*} \frac{\sum_{i\leq n} a_i}{n}-\frac{\sum_{i\leq n+1} a_i}{n+1}&\geq\frac{\sum_{i\leq n} a_i}{n}-\frac{\sum_{i\leq n} a_i + \left(\frac{\sum_{i\leq n} a_i}{n}-\frac{\sum_{i<j\leq n}(a_i-a_j)^2}{2\sum_{i\leq n} a_i}\right)}{n+1}\\ &= \frac{\sum_{i<j\leq n}(a_i-a_j)^2}{2(n+1)\left(\sum_{i\leq n}a_i\right)}\\ &\geq \frac{\sum_{i<j\leq n}(a_i-a_j)^2}{2(n+1)n}\qquad\text{since all }a_i\leq 1\\ &\geq \frac{(n-k)\cdot k\cdot \left(\frac d{k+1}\right)^2}{2(n+1)n}\qquad\text{since for }j\geq k+1,a_j\leq 1-\frac d{k+1}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\text{by Lemma 1 and } a_1=\dots=a_k=1 \end{align*} But note that $k,d$ are constants, therefore the top is linear in $n$ while the bottom is quadratic in $n$. Therefore, we have that \[S(N)=\sum_{n=k+1}^N \frac{(n-k)\cdot k\cdot \left(\frac d{k+1}\right)^2}{2(n+1)n}\]is unbounded as $N$ increases without bound (by limit comparison test to the harmonic series with finitely many initial terms chopped off). But we also have \begin{align*} 1-\frac d{k+1}&=\frac{\sum_{i\leq k+1} a_i}{k+1}\\ &\geq \frac{\sum_{i\leq k+1} a_i}{k+1}-\frac{\sum_{i\leq N+1}a_i}{N+1}\\ &=\sum_{n=k+1}^N\frac{\sum_{i\leq n} a_i}{n}-\frac{\sum_{i\leq n+1} a_i}{n+1}\\ &\geq\sum_{n=k+1}^N \frac{(n-k)\cdot k\cdot \left(\frac d{k+1}\right)^2}{2(n+1)n}\\ &=S(N), \end{align*}which is a contradiction for very large $N$. Thus, we must have that $a_i=1$ for all $i$, as desired. $\square$ Second solution, by eisirrational: Let $$AM = \frac{a_1+\cdots+a_n}{n}, \qquad QM = \sqrt{\frac{a_1^2+\cdots+a_n^2}{n}}, \qquad \overline{AM}=\frac{a_1+\cdots+a_{n+1}}{n+1}, \qquad \overline{QM}=\sqrt{\frac{a_1^2+\cdots+a_{n+1}^2}{n+1}}$$We are given that $\overline{QM}\leqslant AM$. Notice that by Power Mean, $QM\geqslant AM$ and $\overline{QM}\geqslant \overline{AM}$, with equality holding if and only if the sequence is constant. Claim 1. $a_{n+1}^2 \leqslant AM^2-n(QM^2-AM^2) = (n+1)AM^2 - n QM^2$. Proof. We have that $AM^2 \geqslant \overline{QM}^2$. Now notice that $$\overline{QM}^2=\frac{1}{n+1} \left(a_1^2+\cdots+a_{n+1}^2\right)=\frac{1}{n+1}(nQM^2+a_{n+1}^2)$$The result follows. Claim 2. $\overline{QM}^2-\overline{AM}^2 \geqslant \frac{n}{n+1} (QM^2-AM^2)$. Proof. We have $$\overline{QM}^2=\frac{nQM^2+a_{n+1}^2}{n+1}$$and $$\overline{AM}^2=\frac{a_{n+1}^2+2a_{n+1}nAM+n^2AM^2}{(n+1)^2}$$so thus we have $$\overline{QM}^2-\overline{AM}^2=\frac{n(n+1)QM^2-n^2AM^2+na_{n+1}^2-2a_{n+1}n(AM)}{(n+1)^2}$$Now let $x=AM^2$ and $y=QM^2-AM^2$; we know that $y\geqslant 0$. The above equation becomes $$\overline{QM}^2-\overline{AM}^2=\frac{nx+n(n+1)y+n(a_{n+1}^2-2a_{n+1}n(AM))}{(n+1)^2}$$Now, since the function $f(z)=z^2-2AMz$ is decreasing for $z\leqslant AM$ and $a_{n+1}^2\leqslant AM^2-n(QM^2-AM^2)\leqslant AM^2$, it follows that $$a_{n+1}^2-2AMa_{n+1}=f(a_{n+1})\geqslant f(\sqrt{(n+1)AM^2-nQM^2})=f(\sqrt{x-ny})=x-ny-2\sqrt{x(x-ny)}\geqslant x-ny-(x+(x-ny))=-x$$Thus, $$\overline{QM}^2-\overline{AM}^2=\frac{nx+n(n+1)y+nf(a_{n+1})}{(n+1)^2}\geqslant \frac{ny}{n+1}$$as desired. Returning to the problem, assume for the sake of contradiction that the sequence $a_1, \cdots $ is not constant. Then, for some index $n$, $a_1, \cdots, a_n$ is not constant. Let $QM_{i}=\sqrt{\tfrac{a_1^2+\cdots+a_i^2}{i}}$ and $AM_{i}=\frac{a_1+\cdots+a_i}{i}$; we know that $QM_n>AM_n$, so let $\alpha=QM_n^2-AM_n^2$. By Claim 2, we have $$QM_{n+1}^2-AM_{n+1}^2 \geqslant \frac{n}{n+1}\alpha, \qquad QM_{n+2}^2-AM_{n+2}^2\geqslant \frac{n+1}{n+2}\cdot \frac{n}{n+1}\alpha=\frac{n}{n+2}\alpha, \qquad \cdots$$so it is easily seen by induction that $QM_{i}^2-AM_i^2\geqslant \frac{n}{i}\alpha$ for all $i\geqslant n$. But now we have $AM_i^2\leqslant AM_n^2 - (QM_{n+1}^2-AM_{n+1}^2) - (QM_{n+2}^2-AM_{n+2}^2)-\cdots-(QM_i^2-AM_i^2) \leqslant AM_n^2-\tfrac{n}{n+1}\alpha-\tfrac{n}{n+2}\alpha-\cdots-\tfrac{n}{i}\alpha$. As the Harmonic Series diverges, for sufficiently large $i$, $AM_n^2<\tfrac{n}{n+1}\alpha+\tfrac{n}{n+2}\alpha+\cdots+\tfrac{n}{i}\alpha$, which implies $AM_i^2<0$, giving a contradiction. $\square$

MOUNT INEQUALITY >

Here is my solution found during the contest. Assume that the sequence is nonconstant. First, let us set up some notations. \begin{align*} S_n &= \frac{a_1+a_2+\hdots+a_n}{n} \\ T_n &= \sqrt{\frac{a_1^2+a_2^2+\hdots+a_n^2}{n}} \\ D_n &= \sum_{1\leq i,j\leq n} (a_i-a_j)^2 \end{align*}Notice that $D_n = n^2(T_n^2-S_n^2)$. Moreover, from the problem's condition and AM-QM inequality, we get $S_n\geq T_{n+1}\geq S_{n+1}$. Thus $a_{n+1}\leq S_n$ and $S_1\geq S_2\geq S_3\geq \hdots$. This means that the sequence $a_1,a_2,\hdots,a_n$ is bounded above by $a_1$. Now we will get more serious and derive a more precise estimate of $a_{n+1}$. Set $\delta<\tfrac{1}{2a_1}$ be a very small positive real. The key claim is the following bound. Claim: $a_{n+1} < S_n - \delta\frac{D_n}{n}$ for any $n$. Proof: From the condition $T_{n+1}\leq S_n$, we get that $$a_{n+1}^2 + (a_1^2+\hdots+a_n^2) \leq (n+1)S_n^2$$or equivalently, $$a_{n+1}^2 \leq (n+1)S_n^2 - nT_n^2 = S_n^2 - \frac{D_n}{n}.$$Hence $$S_{n+1}-a_n\geq \frac{D_n}{n(S_n+a_{n+1})} \geq \frac{\delta D_n}{n}$$as desired. $\blacksquare$ One corollary of this claim is $D_n < \tfrac{nS_n}{\delta} = O(n)$ for all $n$ or $D_n$ has a linear growth. However, from this bound on $a_{n+1}$, we will estimate $D_{n+1}$ and eventually derive a contradiction. Claim: $D_{n+1} > \frac{n+1}{n}\cdot D_n + \frac{\delta^2\cdot D_n^2}{n}$ for any $n$. Proof: Direct computation using the above claim. \begin{align*} D_{n+1} - D_n &= (a_{n+1}-a_1)^2 + (a_{n+1}-a_2)^2 + \hdots + (a_{n+1}-a_n)^2 \\ &= na_{n+1}^2 - 2(a_1+\hdots+a_n)a_{n+1} + (a_1^2+\hdots+a_n^2) \\ &= na_{n+1}^2 - 2nS_n\cdot a_{n+1} + nT_n^2 \\ &= n(S_n-a_{n+1})^2 + n(T_n^2-S_n^2) \\ &> \frac{\delta^2\cdot D_n^2}{n} + \frac{D_n}{n} \end{align*}So we are done. $\blacksquare$ We claim that the above recursion force $D_n$ to grow quadratically. This will contradict to the corollary mentioned above. To see why, set $x_n = \tfrac{D_n}{n}$. Then the second claim becomes $$x_{n+1} > x_n + \delta^2x_n^2\cdot\frac{n}{n+1} > x_n + \frac{\delta^2}{2}x_n^2.$$Thus, since $a_i$ is nonconstant, $x_m>0$ for some $m$. Hence $$x_{n+m} > x_n + m\cdot\frac{\delta^2}{2}x_n^2$$or $x_n$ grows linearly. Hence $D_n$ grows quadratically, so contradiction.

Let $V_n$, $Q_n$ be the average of the first $n$ terms and the average of the squares of the first $n$ terms, respectively, and let $S_n=n(Q_n-V_n^2)$. Note that $S_n \geq 0$ with equality if and only if the first $n$ terms are equal. By expanding both sides, $$S_{n+1}-S_n=\frac{n}{n+1}(a_{n+1}-V_n)^2$$so $S_n$ is nonincreasing. Now, note that the condition implies $$a_{n+1}^2 \leq V_n^2-S_n \leq Q_n-S_n$$$$Q_{n+1}=\frac{1}{n+1}\left(nQ_{n}+a_{n+1}^2\right) \leq Q_{n}-\frac{S_n}{n+1}.$$so if $S_k>0$ for any $k$, we have that for all $N>k$, $$Q_{N} \leq Q_{N-1}-\frac{S_{N-1}}{N}\leq Q_{N-2}-\frac{S_{N-1}}{N}-\frac{S_{N-2}}{N-1} \leq \ldots \leq Q_{k}-\frac{S_{N-1}}{N}-\frac{S_{N-2}}{N-1}- \ldots - \frac{S_k}{k+1} \leq Q_k-S_k\left(\frac{1}{k+1}+\frac{1}{k+2} \ldots +\frac{1}{N}\right)$$but since the harmonic series diverges, this forces $Q_n$ to be negative for large $N$, contradiction. Thus, $S_n$ is always $0$, implying the result.

Define $r_n = \frac{a_1+a_2+\cdots+a_n}n$, $q_n=\sqrt{\frac{a_1^2+a_2^2+\cdots+a_{n}^2}{n}}$. We have $q_i\geq r_i\geq q_{i+1}\geq r_{i+1}\geq 0$ for all $i$, so in particular $r_i$ tends to a limit $L\geq 0$. Now the given condition implies $r_n^2-a_{n+1}^2\geq \frac{\sum_{i<j\leq n}(a_i-a_j)^2}{n}$. Denote $M_n = \max_{i\leq n} a_i$, $m_n = \min_{i\leq n} a_i$. If $\{a_n\}$ is not constant, then for all sufficiently large $n$, there is some $\delta>0$ such that $M_n-m_n\geq \delta$. On the other hand, $\frac{\sum_{i<j\leq n}(a_i-a_j)^2}{n} \geq \frac{(M_n-m_n)^2+(n-2)(M_n-m_n)^2/2}{n} = \frac{(M_n-m_n)^2}2$, so that there is some $D>0$ such that for all sufficiently large $n$, $r_n^2-a_{n+1}^2\geq D$, or $a_{n+1}\leq \sqrt{r_n^2-D}$. Because $r_n$ tends to a limit $L$, the consequence is that there exists some $\epsilon>0$ such that for all sufficiently large $n$, $a_{n+1}\leq L-\epsilon$. But then $r_n = \frac{a_1+a_2+\cdots+a_n}n$ cannot tend to $L$, contradiction.

My solution on the contest: Let $\mu_n=\frac{a_1+a_2+\ldots+a_n}{n}$ be the arithmetic mean of the first $n$ terms of the sequence. Fix an $n$ and let $x_i=a_i-\mu_n$. Suppose $x_m \neq x_{m+1}$ for some $m \in \{1, 2, \ldots, n-1\}$ and let $d=x_{m+1}-x_m$. Note that $a_1^2+\ldots+a_n^2=(x_1+\mu_n)^2+\ldots+(x_n+\mu_n)^2=(x_1^2+x_2^2+\ldots+x_n^2)+n\mu_n^2 \geq x_m^2+x_{m+1}^2+n\mu_n^2 \geq n\mu_n^2+\frac{d^2}{2}$. Then, from the condition it follows, for $n \geq m$, that $\mu_n \geq \sqrt{\mu_{n+1}^2+\frac{d^2}{2(n+1)}}$ or, equivalently, $\mu_n^2-\mu_{n+1}^2 \geq \frac{d^2}{2(n+1)}$. Take $N$ large enough. By summing the inequalities from $m$ to $N$ we get $\mu_m^2 \geq \mu_m^2-\mu_N^2 \geq \frac{d^2}{2}\bigg(\frac{1}{m+1}+\frac{1}{m+2}+\ldots+\frac{1}{N+1} \bigg)$. The right-hand side is the harmonic series, which diverges, while the left-hand side is constant. This is a contradiction. Thus, $a_m=a_{m+1}$ for every $m$. Remark. The objective of my solution was to try and prove a stronger version of the classic QM-AM inequality. Knowing the equality case, it seemed and proved to be wise to express the difference $QM-AM$ (or something analogous) in terms of the discrepancies between the terms and the arithmetic mean.

How many times can you apply QM-AM to solve this problem? The answer: more than 1. Much more than 1. Let $A_n$ and $Q_n$ be the arithmetic and quadratic means. We note that $Q_n\geq A_n\geq Q_{n+1}\geq A_{n+1}$, so both $A_n$ and $Q_n$ are non-increasing. Now, assume that the sequence is non constant. Thus, these sequences are actually decreasing. Now, assume that $a_n\neq a_1$ and consider some $N\geq n$. We can compute \[Q_N^2-A_N^2=\frac{\sum\limits_{1\leq i<j\leq N}(a_i-a_j)^2}{N^2}\geq\frac{\sum\limits_{i=1}^N (a_i-a_n)^2+(a_i-a_1)^2}{N^2}\]If $a_i\in [a_1,a_n]$, applying QM-HM, we see $(a_i-a_n)^2+(a_i-a_1)^2\geq\frac 12(a_n-a_1)^2$. Otherwise, the inequality is immediate (as one of $(a_i-a_n)^2$ and $(a_1-a_i)^2$ is greater than $(a_n-a_1)^2$. Thus, we get that \[Q_N^2\geq A_N^2+\frac{(a_1-a_n)^2}{2N}\geq Q_{N+1}^2+\frac{(a_1-a_n)^2}{2N}\]Thus, we can get that \[Q_n^2\geq Q_N^2+\frac{(a_1-a_n)^2}2\left(\frac 1n+\frac 1{n+1}+\cdots+\frac 1{N-1}\right)= Q_N^2+\frac{(a_1-a_n)^2}2(H_N-H_n)\]where $H_k$ is the $k$th Harmonic Number. As $H_N$ diverges, $Q_n^2$ is infinite, a contradiction.

MassimilianoF wrote: My solution on the contest: Let $\mu_n=\frac{a_1+a_2+\ldots+a_n}{n}$ be the arithmetic mean of the first $n$ terms of the sequence. Fix an $n$ and let $x_i=a_i-\mu_n$. Suppose $x_m \neq x_{m+1}$ for some $m \in \{1, 2, \ldots, n-1\}$ and let $d=x_{m+1}-x_m$. Note that $a_1^2+\ldots+a_n^2=(x_1+\mu_n)^2+\ldots+(x_n+\mu_n)^2=(x_1^2+x_2^2+\ldots+x_n^2)+n\mu_n^2 \geq x_m^2+x_{m+1}^2+n\mu_n^2 \geq n\mu_n^2+\frac{d^2}{2}$. Then, from the condition it follows, for $n \geq m$, that $\mu_n \geq \sqrt{\mu_{n+1}^2+\frac{d^2}{2(n+1)}}$ or, equivalently, $\mu_n^2-\mu_{n+1}^2 \geq \frac{d^2}{2(n+1)}$. Take $N$ large enough. By summing the inequalities from $m$ to $N$ we get $\mu_m^2 \geq \mu_m^2-\mu_N^2 \geq \frac{d^2}{2}\bigg(\frac{1}{m+1}+\frac{1}{m+2}+\ldots+\frac{1}{N+1} \bigg)$. The right-hand side is the harmonic series, which diverges, while the left-hand side is constant. This is a contradiction. Thus, $a_m=a_{m+1}$ for every $m$. Remark. The objective of my solution was to try and prove a stronger version of the classic QM-AM inequality. Knowing the equality case, it seemed and proved to be wise to express the difference $QM-AM$ (or something analogous) in terms of the discrepancies between the terms and the arithmetic mean. What a brillant guy

Let $b_n=\frac{a_1+\dots+a_n}{n}$ and $c_n=\sqrt{\frac{a_1^2+\dots+a_n^2}{n}}$. As has been noted before, we have $b_n \ge c_{n+1} \ge b_{n+1}$ and hence both $b_n$ and $c_n$ converge to the same limit $L$ by monotonicity. As noted in #6, squaring the inequality in the problem leads to \[b_n^2-a_{n+1}^2 \ge \frac{\sum_{i<j \le n} (a_i-a_j)^2}{n} \ge \frac{1}{n}\sum_{i=1}^n (a_i-a_k)^2=a_k^2-2b_na_k+c_n^2\]where $k$ is any fixed index. So \[a_{n+1}^2 \le 2b_n^2-c_n^2-(a_k-b_n)^2.\]So if $a_k \ne L$, letting $n \to \infty$ we have $a_{n+1}<L-\delta$ for all large $n$ and some positive constant $\delta$. But then of course $b_n<L-\frac{\delta}{2}$ (say) for all large $n$ contradicting our assumption that $b_n \to L$.

Let's introduce some notations first. We will work with $x_n$ instead of $a_n$. Let $a_n$ and $q_n$ denote the am and qm of the first $n$ $x_i$'s. The given condition reads $a_n \geq q_{n+1}$ We will first show that $a_n \geq x_{n+1}$. Note that $$a_n \geq q_{n+1} \geq a_{n+1} = \frac {n}{n+1}a_n+\frac {x_{n+1}}{n+1}$$ This rearranges to give the desired result. $\square$ Define the quantity $\delta_n = q_n^2-a_n^2\geq 0$. We know prove an estimate on $\delta_n$. Claim : $ \delta_{n+1}\geq \frac {n}{n+1} \delta_n$ Proof : Note that we have : $$ q_{n+1}^2= \frac { nq_n^2+x_{n+1}^2 }{n+1} \quad a_{n+1}^2 = \frac { x_{n+1}^2 + 2nx_{n+1}a_n + n^2a_n^2 }{(n+1)^2}$$ We have : $$ \delta_{n+1} = q_{n+1}^2-a_{n+1}^2$$$$ \iff \delta_{n+1} = \frac {n(n+1)q_n^2-n^2a_n^2+n (x_{n+1})(x_{n+1}-2a_n)}{(n+1)^2}$$ $$ \iff \delta_{n+1} = \frac {n(n+1)\delta_n+na_n^2+n (x_{n+1})(x_{n+1}-2a_n)}{(n+1)^2}$$ Note that the quadratic function $f(x)=x(x-2a_n)$ attains its minima at $x=a_n$. Also we have $x_{n+1}\leq a_n$ so $f(x_{n+1}) \geq f(a_n)=-a_n^2$. This gives : $$ \iff \delta_{n+1} = \frac {n(n+1)\delta_n+na_n^2+n (x_{n+1})(x_{n+1}-2a_n)}{(n+1)^2} \geq \frac {n(n+1)\delta_n^2+na_n^2+n (-a_n^2)}{(n+1)^2} = \frac {n}{n+1}\delta_n$$ The claim holds. $\square$ Assume that $a_n$ is such that it is not equal to the terms before it. This means we get $x=\delta_n >0$. Pick a large integer $N > n^{100000}$. We have $\delta_{i+n} \geq \frac {nx}{n+i}$ by induction or whatever. Summing up from $i=n$ to $i=N+n$ we have: $$q_n^2+ \underbrace {q_{n+1}^2-a_n^2}_{\leq 0}+ \dots + \underbrace{q_{n+N}^2-a_{n+N-1}^2}_{\leq 0}-a_{N+n}^2 \geq \sum_{i=1}^N \frac {nx}{n+i} $$ This gives : $$ -a_{N+n}^2 \geq \sum_{i=1}^N \frac {nx}{n+i}-q_n^2 +\quad \text {non negative stuff}$$ Since the harmonic sum diverges, eventually $-a_n^2>0$ . We have the desired contradiction. $\blacksquare$

For brevity let $A_n$ and $Q_n$ denote the arithmetic and quadratic means of the first $n$ $a_i$. The given condition (and QM-AM) implies $$Q_1^2 \geq A_1^2 \geq Q_2^2 \geq A_2^2 \geq \cdots,$$and all of these terms are nonnegative. Suppose that $a_1,a_2,\ldots$ is nonconstant and pick $p \neq q$ such that $|a_p-a_q|=d>0$. Then for $n \geq \max(p,q)$, we have \begin{align*}Q_n^2-A_n^2&=\frac{1}{n^2}\left(n\sum_{i=1}^n x_i^2-\sum_{i=1}^n x_i^2-2\sum_{1 \leq i<j \leq n} x_ix_j\right)\\ &=\frac{1}{n^2}\left((n-1)\sum_{i=1}^n x_i^2-2\sum_{1 \leq i<j\leq n}x_ix_j\right)\\ &=\frac{1}{n^2}\left(\sum_{1 \leq i<j \leq n} (x_i-x_j)^2\right)\\ &\geq \frac{1}{n^2}\left((x_p-x_q)^2+\sum_{\substack{1 \leq i \leq n\\ i \not \in \{p,q\}}} ((x_i-x_p)^2+(x_i-x_q)^2)\right)\\ &\geq \frac{1}{n^2}\left(d^2+(n-2)\frac{d^2}{2}\right)\\ &=\frac{d^2}{2n} \end{align*}But now $\sum_{i=1}^\infty (Q_i^2-A_i^2)\geq \frac{d^2}{2}\sum_{i=\max\{p,q\}}^\infty \frac{1}{n}=+\infty$ diverges, hence the aforementioned sequence goes to $-\infty$: absurd. $\blacksquare$ Remark: The key step here is realizing (perhaps intuitively) that if there exists some $|a_p-a_q|=d>0$ then we should be able to get some bound on $Q_n^2-A_n^2$, and then carrying this out and getting an $n^{-1}$-order bound. If one doesn't spot the factorization into the sum of $(x_i-x_j)^2$, it is also possible to perform the following more convoluted (but ultimately isomorphic) process. First, note that $$(n-1)\sum_{i=1}^n x_i^2-2\sum_{1 \leq i<j\leq n} x_ix_j=\left((n-1)(x_1^2+x_2^2)+2\sum_{i=3}^n x_i^2-2x_1x_2-2\sum_{i=3}^n (x_1+x_2)x_i\right)+\underbrace{\left((n-3)\sum_{i=3}^n x_i^2-2\sum_{3 \leq i<j \leq n} x_ix_j\right)}_{\geq 0}.$$Note that subject to $x_3+\cdots+x_n$ fixed, $2\sum_{i=3}^n x_i^2-2\sum_{i=3}^n (x_1+x_2)x_i$ is minimized when $x_3=\cdots=x_n:=x$, so we can bound $$(n-1)(x_1^2+x_2^2)-2x_1x_2+2(n-2)x^2-2(n-2)(x_1+x_2)x.$$This is an upwards-opening quadratic in $x$ whose minimum occurs when $x=\frac{x_1+x_2}{2}$, at which point it equals $$(n-1)(x_1^2+x_2^2)-2x_1x_2-\frac{n-2}{2}(x_1+x_2)^2=\frac{n}{2}(x_1-x_2)^2,$$from which we get the same lower bound of $\frac{d^2}{2n}$.