How to show these two random variables are not equal with positive probability? Given iid $X_1, \dots, X_n \sim N(a, a^2), a \neq 0$,
how can we show that it is not true $\frac{(\sum_i X_i)^2}{n(n+1)} = \frac{\sum_i X_i^2}{2n}$ a.e.?
My thought was checking the moments. I got that
$\mathrm E \frac{(\sum_i X_i)^2}{n(n+1)} =  \mathrm E \frac{\sum_i X_i^2}{2n} = a^2$. But checking  $\mathrm Var [\frac{(\sum_i X_i)^2}{n(n+1)} - \frac{\sum_i X_i^2}{2n}]$ is very complicated, which I think not worth to try?
 A: I believe this kind of thing shouldn't depends on $n$ (unless $n=1$, of course).
So you could try with only two variables and to argue via induction in $n$.
For two random variables you can verify theirs differences. Taking $n=2$ you will have
$$\frac{X_1^2+X^2}{4}-\frac{X_1^2+2X_1X_2+X_2^2}{6} = \frac{X_1^2+X_2^2-4X_1X_2}{12}=\frac{(X_1-X_2)^2-2X_1X_2}{12}$$
Verifying one part
$$P(\frac{(X_1-X_2)^2-2X_1X_2}{12} = 0) = P((X_1-X_2)^2=2X_1X_2)  $$
This could be true only if the r.v's $X_1$ and $X_2$ have the same sign. But you can use the fact they are independent of each other to show that the probability of $X_1$ and $X_2$ to have the same signal is strictly less than one (in fact you can determine this probability explicitly once you know theirs distribution).
The induction argument should be the same, just write the sum $\sum_i^nX_i$ as $\sum_i^{n-1}X_i+X_n$  and now you have reduced to the two variables case.
It's not an elegant solution... But it is what came to mind at first place...

Edit: 
I said that the proof would be by induction, but I was wrong... What I should have said it is seems to me that the same argument should work to general case.
$$\frac{n\sum_{j=1}^nX_j^2-2(\sum_{j=1}^nX_j)^2}{2n(n+1)} = \frac{(n-1)\sum_{j=1}^nX^2_j-4\sum X_iX_j}{2n(n+1)}$$
For $n = 4$
Now you can regroup the terms on the numerator,
$$2(X_1-X_2)^2+2(X_3-X4)^2+X_1^2+X_2^2+X_3^2+X_4^2-4(X_1+X_2)(X_3+X_4)$$ and then you would have 
$$P(2(X_1-X_2)^2+2(X_3-X4)^2+X_1^2+X_2^2+X_3^2+X_4^2=4(X_1+X_2)(X_3+X_4))$$
Now we use the independence of the variables and theirs distribution...
A: If $n=1$ the conclusion does not hold. Asume that $n\geqslant2$ and consider the random variables $$S_n=\sum\limits_{k=2}^nX_k,\qquad T_n=(n+1)\sum\limits_{k=2}^nX_k^2-2S_n^2,$$ then some simple algebraic manipulations show that the event of interest is $$A_n=[(n-1)X_1^2-2S_nX_1+T_n=0],$$ that is, $$A_n=[X_1=U_n]\cup[X_1=V_n],$$ where the random variables $U_n$ and $V_n$ (the roots of the quadratic polynomial in $X_1$) are measurable with respect to $(X_k)_{2\leqslant k\leqslant n}$, in particular, they are independent of $X_1$. Here is a useful result:

Lemma: Let $X$ and $Y$ be independent, $X$ atomless, then $P(X=Y)=0$.

Since the distribution of $X_1$ is continuous, the lemma applied to $X=X_1$ and to $Y=U_n$ and $Y=V_n$ implies that $P(X_1=U_n)=P(X_1=V_n)=0$, hence, not only $P(A_n)\ne1$ but actually $P(A_n)=0$.
Proof of the lemma: By definition and by independence of $X$ and $Y$, $$P(X=Y)=\int\left(\int\mathbf 1_{x=y}\mathrm dP_X(x)\right)\mathrm dP_Y(y)=\int P(X=y)\mathrm dP_Y(y)=0,$$ since $P(X=y)=0$ for every $y$, QED.
