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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

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closed as off-topic by Saad, Cesareo, GNUSupporter 8964民主女神 地下教會, Paul Frost, Eevee Trainer Mar 9 at 0:34

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$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$ For $n$ variables there are two versions.

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

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 $a_{n+1}=a_1$.

  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 $$\sum_{1\leq k<m\leq n}(a_k-a_m)^2\geq0.$$

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for $a,b,c\in R(a\neq b\neq c)$

$(a-b)^2+(b-c)^2+(c-a)^2>0$

$a^2+b^2+c^2>ab+bc+ca$

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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.

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