Sum of cross terms vs sum of squares? What do we know about sum of squares vs sum of cross terms? Does one always dominate the other? Any theorems on that?
e.g
for  
$a^2 + b^2 + c^2 \ < ? > \ ab + ac + bc $
for any number of terms.
Thank you
 A: In general, let
$$f_n(x_1,\ldots,x_n)=\sum_{i,j=1\atop i< j}^nx_ix_j
=\frac12\sum_{i,j=1\atop i\ne j}^nx_ix_j.$$
For each $i\ne j$, $x_ix_j\le\frac12(x_i^2+x_j^2)$ (AM/GM)
with equality iff $x_i=x_j$. Thus
$$f_n(x_1,\ldots,x_n)\le\frac{n-1}2\sum_{i=1}^nx_i^2$$
with equality iff $x_1=x_2=\cdots=x_n$. From this argument,
the constant $\frac12(n-1)$ is best possible.
A: for $a,b,c\in R(a\neq b\neq c)$
$(a-b)^2+(b-c)^2+(c-a)^2>0$
$a^2+b^2+c^2>ab+bc+ca$
A: $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$
For $n$ variables there are two versions.

  
*
  
*$$a_1^2+a_2^2+...+a_n^2\geq a_1a_2+a_2a_3+...+a_na_1.$$

Proof:
$$\sum_{k=1}^n(a_k^2-a_ka_{k+1})=\frac{1}{2}\sum_{k=1}^n(a_k^2-2a_ka_{k+1}+a_{k+1}^2)=\frac{1}{2}\sum_{k=1}^n(a_k-a_{k+1})^2\geq0.$$
Here $a_{n+1}=a_1$.


  
*$$(n-1)(a_1^2+a_2^2+...+a_n^2)\geq2(a_1a_2+a_1a_3+...+a_1a_n+a_2a_3+...+a_{n-1}a_n).$$

Proof:
We need to prove that:
$$(n-1)\sum_{k=1}^na_k^2\geq2\sum_{1\leq k<m\leq n}a_ka_m$$ or
$$\sum_{1\leq k<m\leq n}(a_k-a_m)^2\geq0.$$
