# Is there Inequality relating the arithmetic mean (AM) and the dot product

Is there an inequality of the kind:

$$\sum_{i=1}^n a_{i}b_{i} \geq C_{1}\left(\frac{\sum_{i=1}^{n} a_{i}}{n}\right)C_{2}\left(\frac{\sum_{i=1}^{n} b_{i}}{n}\right),$$

i.e., one relating the dot (inner) product $$\sum\limits_{i} a_{i}b_{i}$$ and the arithmetic means for $$A$$ and $$B$$, where the dot product is the greater quantity?

Here, $$C_1$$ and $$C_2$$ might not necessarily be constants, but possibly involving the maximum or minimum of $$A$$ and/or $$B$$ or something else.

Well-known inequalities like the Rearrangement inequality and the Holder's inequality do not give what I want (neither the Holder’s reverse inequality).

Edit after the comment of @Jack D'Aurizio: the dot product is not $$0$$.

• In Chebyshev's inequality the numbers in $A$ and $B$ are ordered. I need something in the general case.
– sdd
Jul 16, 2021 at 7:23
• I got for $1\geq a,b,c,d>0$ $$ab+cd-\frac{\left|\left(a-b\right)\left(c-d\right)\left(b-c\right)\left(a-d\right)\right|\left(a+c\right)\left(d+b\right)}{4}\geq 0$$.Can you confirm numerically (I use Desmos not Mathematica)?(+1)For the interesting question . Jul 16, 2021 at 11:59
• You need additional contraints, since the dot product can be zero while $\sum a_n,\sum b_n$ can be positive. For instance $\vec{a}=(1,0),\vec{b}=(0,1)$. Jul 16, 2021 at 12:13
• for $a,b,c,d>0$ we have $ac+bd\ \frac{\frac{\sqrt{\left|\left(a-b\right)\left(b-c\right)\left(c-d\right)\left(a-d\right)\left(a-c\right)\left(b-d\right)\right|}}{2\left(a+b+c+d\right)^{2}}\left(a+c\right)\left(b+d\right)}{4}\geq0$ Jul 16, 2021 at 12:20
• @sdd can you confirm ?Thanks . Jul 16, 2021 at 12:48

If $$0 < m_1 \le a_i \le M_1$$ and $$0 < m_2 \le b_i \le M_2$$ for all $$i$$, then $$\sum_{i=1}^n a_ib_i \ge n\left(1 - \frac{(M_1 - m_1)(M_2 - m_2)}{4\sqrt{M_1m_1M_2m_2}}\right)\left(\frac{1}{n}\sum_{i=1}^n a_i\right)\left(\frac{1}{n}\sum_{i=1}^n b_i\right).$$