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 $0$.

  • $\begingroup$ In Chebyshev's inequality the numbers in $A$ and $B$ are ordered. I need something in the general case. $\endgroup$
    – sdd
    Jul 16, 2021 at 7:23
  • $\begingroup$ 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 . $\endgroup$
    – Erik Satie
    Jul 16, 2021 at 11:59
  • 4
    $\begingroup$ You need additional contraints, since the dot product can be zero while $\sum a_n,\sum b_n$ can be positive. For instance $\vec{a}=(1,0),\vec{b}=(0,1)$. $\endgroup$ Jul 16, 2021 at 12:13
  • $\begingroup$ for $a,b,c,d>0$ we have $ac+bd\ \frac{\frac{\sqrt{\left|\left(a-b\right)\left(b-c\right)\left(c-d\right)\left(a-d\right)\left(a-c\right)\left(b-d\right)\right|}}{2\left(a+b+c+d\right)^{2}}\left(a+c\right)\left(b+d\right)}{4}\geq0$ $\endgroup$
    – Erik Satie
    Jul 16, 2021 at 12:20
  • $\begingroup$ @sdd can you confirm ?Thanks . $\endgroup$
    – Erik Satie
    Jul 16, 2021 at 12:48

1 Answer 1


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

[1] Dragomir and Khan, Two Discrete Inequalities of Grüss Type Via Pólya-Szegö and Shisha Results for Real Numbers.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.