2
$\begingroup$

Let $(a_k)_1^n$ and $(b_k)_1^n$ be two real sequences, and suppose that $(b_k)$ is nonnegative and decreasing. For $k \in \{1,2,\ldots,n\}$, define $S_k = \sum_{i=1}^k a_i$. Let $M = \max \{S_1,\ldots, S_n\}$ and $m = \min \{S_1,\ldots,S_n\}$. Prove that $$mb_1 \le \sum_{i=1}^n a_i b_i \le Mb_1$$

I am trying to prove this by assuming that $m = S_j$, so I get $$(a_1 + a_2 + \cdots + a_j)b_1 \le a_1b_1 + a_2b_2 + \cdots + a_nb_n $$ which is equivalent to \begin{align} & a_2(b_1-b_2) + \cdots + a_j(b_1-b_j) \\ \le {} & a_{j+1}b_{j+1} + \cdots + a_nb_n \tag 1 \end{align} Now if the $a_i$s are all positive then it is easy since from the fact that $(b_k)$ is a decreasing sequence, $$a_2(b_1-b_2) + \cdots + a_j(b_1-b_j) \le (a_2 + \cdots +a_j)(b_1-b_j)$$ But we also know that $S_1 \ge S_j$, so $a_2 + \cdots + a_j \le 0$, which implies that the LHS of $(1)$ is less than or equal to $0$, and as $S_j\le S_n \Rightarrow a_{j+1} + \cdots + a_n \ge 0$, so $$a_{j+1}b_{j+1} + \cdots + a_nb_n \ge (a_{j+1} + \cdots + a_n)b_n \ge 0$$ therefore the RHS of $(1)$ is greater than or equal to $0$, hence $(1)$ is proven. In a similar way, we can also prove that $\sum_{i=1}^n a_ib_i \le Mb_1$.

But the problem is that $(a_k)$ is a real sequence, not a positive sequence, which means there may be negative terms in it, and I don't have any idea how to proceed. Can anyone help? Thanks.

$\endgroup$
2
  • $\begingroup$ Shouldn't the lower bound be $b_nm$? $\endgroup$
    – SL_MathGuy
    Dec 25, 2019 at 4:14
  • $\begingroup$ Well, the original question stated the lower bound as $mb_1$, not $mb_n$... $\endgroup$
    – Vann
    Dec 25, 2019 at 4:21

2 Answers 2

2
$\begingroup$

Since $a_1 = S_1$ and $a_i = S_i - S_{i-1}, \ i = 2, 3, \cdots, n$, we have \begin{align} &\sum_{i=1}^n a_i b_i \\ =\ & S_1b_1 + (S_2-S_1)b_2 + (S_3-S_2)b_3 + (S_4 - S_3)b_4 + \cdots + (S_n - S_{n-1})b_n\\ =\ &S_1(b_1 - b_2) + S_2(b_2 - b_3) + S_3(b_3 - b_4) + \cdots + S_{n-1}(b_{n-1} - b_n) + S_nb_n. \end{align} Since $b_n \ge 0$ and $b_i - b_{i+1} \ge 0, \ i=1, 2, \cdots, n-1$ and $m \le S_i \le M, \ i = 1, 2, \cdots, n$, we have \begin{align} &\sum_{i=1}^n a_i b_i \\ \ge\ & m(b_1 - b_2) + m(b_2 - b_3) + m(b_3 - b_4) + \cdots + m(b_{n-1} - b_n) + mb_n\\ =\ & mb_1 \end{align} and \begin{align} &\sum_{i=1}^n a_i b_i \\ \le\ & M(b_1 - b_2) + M(b_2 - b_3) + M(b_3 - b_4) + \cdots + M(b_{n-1} - b_n) + Mb_n\\ \le\ & Mb_1. \end{align} We are done.

See: Abel's summation by parts

$\endgroup$
1
$\begingroup$

______Editing my previous comment & extending the answer_____________

Case 1: When {$a_i$} is a sequence of positive terms.

Upper bound can be obtained easily.

Since {$b_i$} is decreasing, $b_1\geq b_i$ for all $i=1,2...n$. Hence, $\sum_{i=1}^{n}a_ib_i\leq \sum_{i=1}^{n}a_ib_1=b_1\sum_{i=1}^{n}a_i=b_1S_n\leq b_1M$.

Similarly, for the lower bound,

$\sum_{i=1}^{n}a_ib_i\geq b_1a_1=b_1m$, since $a_i>0$ for all $i$ & $S_1=a_1<S_i$ for all i.

Case 2: When {$a_i$} is a sequence of negative terms

$\sum_{i=1}^{n}a_ib_i\leq a_1b_1$. Note that $S_1=a_1$.Since $a_i<0$ & $S_1 \geq S_i$ for all $i$, $M=S_1$ so that $\sum_{i=1}^{n}a_ib_i\leq Mb_1$.

Similarly, for the lower bound,

$S_n \leq S_i$ for all $i$ & $b_1 \geq b_i$ for all $i$. So $m=S_n$, $\sum_{i=1}^{n}a_ib_i \geq b_1 \sum_{i=1}^{n}a_i=b_1S_n=b_1m$.

Case 3: When {$a_i$} consists of + & - terms,

Observe that $\sum -|a_i|b_i \leq \sum a_ib_i \leq \sum |a_i|b_1$

Rightmost term is the one we considered under case1 & leftmost term is the one under case 2. Since the result was proved for both cases, it trivially holds for this case as well.

$\endgroup$
6
  • $\begingroup$ hmm..but how could you know that $\sum_{i=1}^{n} a_ib_i \le \sum_{i=1}^{n} a_ib_1$? Because the $a_i$s can also be negative(?) $\endgroup$
    – Vann
    Dec 25, 2019 at 4:31
  • $\begingroup$ It doesn't matter. $\sum a_ib_i= a_1b_1+a_2b_2+...+a_nb_n \leq a_1b_1+a_2b_1+...+a_nb_1=b_1(a_1+a_2+...+a_n) $ $\endgroup$
    – SL_MathGuy
    Dec 25, 2019 at 4:34
  • $\begingroup$ What if we take $a_1 = -3, a_2 = -2, a_3 = -1, b_1 = 3, b_2 = 2, b_1 = 1$, then $\sum_{i=1}^{3} a_ib_i = -14$, however $\sum_{i=1}^{n} a_ib_1 = -18$, a contradiction? But the inequality $\sum_{i=1}^{n} a_ib_i \le Mb_1$ is still true though ($-14 \le (-3)(3)$). $\endgroup$
    – Vann
    Dec 25, 2019 at 4:52
  • $\begingroup$ Yeah correct. I missed that point. Editing my answer to indicate the case where {$a_i$} is a sequence of positive terms.But, according to your example, $m=-6$ & $b_1=1$ so that $mb_1=-6$. In this case, it can't be true that $mb_1 \leq \sum a_ib_i=-18$ . $\endgroup$
    – SL_MathGuy
    Dec 25, 2019 at 5:16
  • $\begingroup$ Ups sorry...I wrote $b_1 = 3, b_2 = 2$, and $b_1 = 1$ again in my previous comment, where the latter should be $b_3$ instead of $b_1$...so $mb_1 = -18 \le \sum a_ib_i = -18$, which is still true I guess. $\endgroup$
    – Vann
    Dec 25, 2019 at 5:24

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.