Assume real numbers $a_i,b_i\,(i=0,1,\cdots,2n)$ satisfy the following conditions: (1) for $i=0,1,\cdots,2n-1$, we have $a_i+a_{i+1}\geq 0$; (2) for $j=0,1,\cdots,n-1$, we have $a_{2j+1}\leq 0$; (2) for any integer $p,q$, $0\leq p\leq q\leq n$, we have $\sum_{k=2p}^{2q}b_k>0$. Prove that $\sum_{i=0}^{2n}(-1)^i a_i b_i\geq 0$, and determine when the equality holds.
Problem
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Tags: induction, strong induction, inequalities unsolved, inequalities
math154
20.09.2011 08:10
Let $c_i=(-1)^i a_i\ge 0$ for $0\le i\le 2n$, so $c_0\ge c_1\le c_2\ge c_3\le \cdots \ge c_{2n-1} \le c_{2n}$ (*). We show that $\sum_{i=0}^{2n}c_ib_i\ge 0$ (with equality only for $c_i=0$ for all $0\le i\le 2n$) by strong induction on $n\ge0$. The base case is trivial, so suppose that $n\ge 1$ and the result is true for $0,\ldots, n-1$. Then by (*), there exists $0\le r\le n-1$ such that $c_i\ge c_{2r+1}$ for all $0\le i\le 2n$, whence \[\sum_{i=0}^{2n}c_ib_i = c_{2r+1}\sum_{k=0}^{2n}b_k + \sum_{i=0}^{2r}(c_i-c_{2r+1})b_i + \sum_{i=2r+2}^{2n}(c_i-c_{2r+1})b_i \ge 0\]by the inductive hypothesis and condition (3) for $(p,q)=(0,n)$, with equality iff $c_{2r+1}=0$ and $c_i-c_{2r+1}=0$ for all $0\le i\le 2n$, as desired.
MathPanda1
17.08.2015 06:32
Why is $c_{2r+1}\sum_{k=0}^{2n}b_k + \sum_{i=0}^{2r}(c_i-c_{2r+1})b_i + \sum_{i=2r+2}^{2n}(c_i-c_{2r+1})b_i \ge 0$ true? Thanks!
Complex2Liu
30.04.2016 18:10
Use Abel equality together with induction.
We define $c_i=(-1)^i a_i$ as same as #2, and $B_i\stackrel{\text{def}}{=}\sum\nolimits_{i=2n-i+1}^{2n} b_i.$ I will use induction on $n$ to claim that $\sum_{i=0}^{2n} c_ib_i\ge 0.$ It's evident that $B_{2i+1}$ is always positive for $i=0,2,\dots,n.$
Clearly if $c_{2n}b_{2n}+c_{2n-1}b_{2n-1}\ge 0$ then we are done by our induction hypothesis, so it must be $\mid b_{2n}\mid <\mid b_{2n-1}\mid,$ i.e. $B_2=b_{2n}+b_{2n-1}<0.$
Now consider $c_{2n}b_{2n}+c_{2n-1}b_{2n-1}+c_{2n-2}b_{2n-2}+c_{2n-3}b_{2n-3},$ Applying Abel equality yields
\[(c_{2n}-c_{2n-1})B_1+(c_{2n-1}-c_{2n-2})B_2+(c_{2n-2}-c_{2n-3})B_3+c_{2n-3}B_4<0,\]which amounts to $B_4<0,$ this is because we have already obtained $B_2<0,$ so only $c_{2n-3}B_4$ could be negative.
Furthermore we yields that $B_{2i}<0$ for $i=1,2,\dots,n.$ Therefore
\[\sum_{i=0}^{2n}c_ib_i=\sum_{i=0}^{2n-1}(c_{2n-i}-c_{2n-1-i})B_{i+1}+c_{0}B_{2n+1}\ge 0.\]as desired. The equality holds for $a_0=a_1=\dots=a_{2n}=0.$ $\square$
guptaamitu1
19.12.2021 14:22
In this handout (P7), the condition $\sum_{k=2p}^{2q}b_k>0$ was replaced by $\sum_{k=2p}^{2q}b_k \ge 0$. So can someone please tell all the equality cases here, since there are also non-trivial equality cases (for example, here $a_0 = a_1 = a_2$ and $(b_0,b_1,b_2) = (1,-2,1)$.