Vtyjkyuk

Given a finite increasing sequence of real numbers $x_i_i=1^n$ consisting of at least two elements, how can we show that
$$
frac1nsum_i=1^nx_ix_n-i+1 < left(frac1nsum_i=1^nx_iright)^2
$$
without expanding. I attempted applying Holder's inequality but failed. Maybe I missed something.

Hints:

1) the sum in LHS is by Rearrangement Inequality the lowest possible of form $x_ix_pi(i)$ where $pi$ is a permutation.

2) unless the $x_i$ are strictly increasing somewhere, equality is possible.

P.S. For some more detail, note that $nsum x_ix_n−i+1<left(sum x_iright)^2$ follows from $n$ rearrangements summed together. Each such rearrangement has an RHS of form $sum x_i x_i+j$ where $j=0,1,...n−1$, and the sum in LHS is lower than each of them.

This should be wrong.
For example for $x_i = 1$ this is an equality isn't it?
$$ frac1n sum limits_i=1^n 1^2 = 1 = 1^2 = (frac1n n)^2 = (frac1n sum limits_i=1^n 1)^2 $$

Let
$$
barx=frac1nsum_i=1^nx_itag1
$$
If $x_ile x_i+1$ and $x_1lt x_n$, we can find a $k$ so that $x_klebarxlt x_k+1$. Then
$$
barx-x_n-i+1ge0iff ige n-k+1tag2
$$
Therefore,
$$
beginalign
left(frac1nsum_i=1^nx_iright)^2-frac1nsum_i=1^nx_ix_n-i+1
&=frac1nsum_i=1^nx_ibarx-frac1nsum_i=1^nx_ix_n-i+1tag3\
&=frac1nsum_i=1^nx_ileft(barx-x_n-i+1right)tag4\
&=frac1nsum_i=1^nleft(x_i-x_n-k+1right)left(barx-x_n-i+1right)tag5
endalign
$$
Explanation:

$(3)$: apply $(1)$

$(4)$: distributive property

$(5)$: subtract $frac1nx_n-k-1sumlimits_i=1^nleft(barx-x_n-i+1right)=0$

Applying $(2)$,
$$
beginalign
ige n-k+1&impliesbarx-x_n-i+1ge0quadtextandquad x_i-x_n-k+1ge0tag6\
ilt n-k+1&impliesbarx-x_n-i+1lt0quadtextandquad x_i-x_n-k+1le0tag7
endalign
$$
Thus, the terms in $(5)$ are non-negative. Furthermore, either the $i=1$ term or the $i=n$ term of $(5)$ is positive; that is, $barx$ cannot equal either $x_1$ or $x_n$, and $x_n-k+1$ cannot equal both $x_1$ and $x_n$.

Summarizing, if $x_ile x_i+1$ and $x_1lt x_n$,
$$
frac1nsum_i=1^nx_ix_n-i+1ltleft(frac1nsum_i=1^nx_iright)^2tag8
$$

$$frac1nsum_i=1^nx_ix_n-i+1 < left(frac1nsum_i=1^nx_iright)^2iffsum_i=1^nx_ix_n-i+1 < frac1nleft(sum_i=1^nx_iright)^2$$

When $n$ is odd one has $LHS=2(x_1x_n+cdots+x_fracn-12x_fracn+32)+x^2_fracn+12$

When $n$ is even one has $LHS=2(x_1x_n+cdots+x_fracn2x_fracn+22)$

In both cases we have trivially that a sum of squares distinct of zero is greater than $0$.

$$0lt n((x_n-x_1)^2+cdots +(x_fracn+32-x_fracn-12)^2)text when n is odd \0lt n((x_n-x_1)^2+cdots +(x_fracn+22-x_fracn2)^2)text when n is even $$