Interesting inequality I'm new to Mathematics Stack Exchange. I have this inequality:
$$\sum_{i=1}^{2013}(x_i-\sqrt{2})(x_i+\sqrt{2}) \geq \sum_{i=1}^{2012}x_ix_{i+1}+x_{2013}x_{1}-3 $$ where $x_{1}, x_{2},...$ are integers all distinct. How to approach it?
 A: Sketch/Hints:  Using $(x-\sqrt{2})(x+\sqrt{2}) = x^2-2$ the left hand side becomes $\displaystyle \sum_{i=0}^{2013} x_i^2 - 4026$.
Then use the fact that $\displaystyle x^2+(x+k)^2-2x(x+k) = k^2$ to transform the given inequality into the equivalent form $$\sum_{i=1}^{2013} \frac{(x_{i+1}-x_i)^2}{2} \geq 4023,$$ where $x_{2014}=x_1$.
Now use the fact that the integers are all distinct to show that $\displaystyle \sum_{i=1}^{2013}|x_{i+1}-x_i| \geq 4024$, with all summands positive integers, and hence equality implying that at least two of the summands are $1$.  This is the trickiest step, so here's a sketch of a proof for this:  use induction on the number of terms, showing that $\displaystyle f_n(\overline{x}) = \sum_{i=1}^n |x_{i+1}-x_i| \geq 2(n-1)$, where $x_{n+1}=x_1$.  The base case $n=2$ is easy.  For larger $n$, choose some $x_i$ which is neither the smallest nor the largest, and remove that $x_i$ from the sequence, then increase every $x_j$ which was less than $x_i$ by $1$, obtaining a reduced sequence of $n-1$ many $y$'s.  By the induction hypothesis, $f_{n-1}(\overline{y}) \geq 2(n-2)$, and it is also true that $f_n(\overline{x}) \geq f_{n-1}(\overline{y})+2$ (consider how the differences between numbers less than $x_i$ and those greater than $x_i$ change when we shift the numbers less than $x_i$ up by $1$).  The inductive step follows.
Finally use the fact that $g(x) = x^2$ is convex to obtain the result in both the case of equality above using the fact that two summands are $1$, and in the case of $\displaystyle \sum_{i=1}^{2013}|x_{i+1}-x_i| \geq 4025$.
A: I thought I would be able to solve the problem, but apparently, I can't. I'll post it because I typed it all already and maybe you can still use is.
We know that all the $x_i$ are different integers. Because we have $n$ ($n=2013$, but this inequality is true for general $n$) of them, there are at least two $x_i$ with difference $n-1$. Suppose w.l.g. that $x_j-x_1\geq n-1$, with $1<j\leq n$. We now know that
\begin{align}
\sum_{i=1}^{j-1}x_{i+1}-x_i&=x_j-x_1\geq n-1\\
\sum_{i=j}^n x_{i}-x_{i+1}&=x_j-x_i\geq n-1
\end{align}
Using the quadratic-arithmetic mean
$$
\sqrt{\frac{\sum a_i^2}n}\geq \frac {\sum a_i}n
$$
with $a_i=\left|x_{i+1}-x_i\right|$, we find that
\begin{align}
\sqrt{\frac{\sum_{i=j}^n (x_{i+1}-x_i)^2}{n-j+1}}&\geq \frac {\sum_{i=j}^n |x_{i+1}-x_j|}{n-j+1}\geq \frac {\sum_{i=j}^nx_{i+1}-x_i}{n-j+1}\geq \frac{(n-1)}{n-j+1}\\
\sqrt{\frac{\sum_{i=1}^{j-1} (x_{i}-x_i+1)^2}{j-1}}&\geq \frac {\sum_{i=1}^{j-1} |x_{i}-x_{i+1}|}{j-1}\geq \frac {\sum_{i=1}^{j-1}x_{i}-x_{i+1}}{j-1}\geq \frac{(n-1)}{j-1}
\end{align}
Squaring both sides, multiplying by $n$ and summing the two inequalities gives
\begin{align}
\sum_{i=1}^n(x_{i+1}-x_i)^2\geq(n-1)^2\left(\frac1{j-1}+\frac1{n-j+1}\right)
\end{align}
With some simple calculus (differentiation) we find that the right hand side is minimal for $j=\frac{n+2}{2}$ with minimum $\frac 4n$. Thus, we get
\begin{align}
\sum_{i=1}^n(x_{i+1}-x_i)^2\geq(n-1)^2\left(\frac1{j-1}+\frac1{n-j+1}\right)\geq 4\frac{(n-1)^2}n
\end{align}
The problem is that the estimation for $j$ is too sharp, so the equality which now remains to be proven isn't always true.
