Let $a_1,a_2,\ldots,a_n$ be non-negative real numbers ,$S_k= \sum\limits_{i=1}^{k}a_i $ $(1\le k\le n)$.Prove that$$\sum\limits_{i=1}^{n}\left(a_iS_i\sum\limits_{j=i}^{n}a^2_j\right)\le \sum\limits_{i=1}^{n}\left(a_iS_i\right)^2$$
Problem
Source: China Mianyang , 16 Aug 2016
Tags: inequalities, algebra
Exista
16.08.2016 15:26
Just Abel(as a verb) LHS.
CJA
19.08.2016 18:14
Exista wrote: Just Abel(as a verb) LHS. ccan you write a solution?thank you
YanYau
20.08.2016 08:11
This was my solution when I took the CWMI paper, probably not the most beautiful, but it took less than half a page for me to write and it was very straight forward: $$\begin{aligned} \sum_{i=1}^{n}\left(a_iS_i\right)^2&=\sum_{i=1}^{n}a_i^2\left(\sum_{j=1}^i a_i\right)^2 \\&=\sum_{i=1}^{n}a_i^2\left(\sum_{j=1}^i a_i^2+ 2\sum_{1\leq j < k \leq i} a_ja_k\right) \\&=\sum_{1\leq j \leq i \leq n} a_j^2a_i^2+2\sum_{1\leq j < k \leq i\leq n}a_ja_ka_i^2 \end{aligned}$$ $$\begin{aligned} \sum_{i=1}^{n}\left(a_iS_i\sum_{j=i}^{n}a^2_j\right)&=\sum_{i=1}^{n}\left(a_i\left(\sum_{k=1}^i a_i\right)\left(\sum_{j=i}^{n}a^2_j\right)\right) \\&=\sum_{i=1}^{n}\left(a_i\sum_{1\leq k \leq i\leq j}^{n}a_ka^2_j\right) \\&=\sum_{1\leq k \leq i\leq j\leq n}^{n}a_ka_ia^2_j \\&=\sum_{1\leq i\leq j\leq n}^{n}a_i^2a^2_j + \sum_{1\leq k < i\leq j\leq n}^{n}a_ka_ia^2_j \end{aligned}$$ $$\begin{aligned} \sum\limits_{i=1}^{n}\left(a_iS_i\sum\limits_{j=i}^{n}a^2_j\right)&\le \sum\limits_{i=1}^{n}\left(a_iS_i\right)^2 \\\sum_{1\leq i\leq j\leq n}^{n}a_i^2a^2_j + \sum_{1\leq k < i\leq j\leq n}^{n}a_ka_ia^2_j &\leq \sum_{1\leq j \leq i \leq n} a_j^2a_i^2+2\sum_{1\leq j < k \leq i\leq n}a_ja_ka_i^2 \\0&\leq \sum_{1\leq j < k \leq i\leq n}a_ja_ka_i^2 \end{aligned}$$ Since $a_i$ is non-negative, this inequality is always true, with equality when $a_i=0$
sqing
20.08.2016 14:53
YanYau wrote: This was my solution when I took the CWMI paper, probably not the most beautiful, but it took less than half a page for me to write and it was very straight forward: $$\begin{aligned} \sum_{i=1}^{n}\left(a_iS_i\right)^2&=\sum_{i=1}^{n}a_i^2\left(\sum_{j=1}^i a_i\right)^2 \\&=\sum_{i=1}^{n}a_i^2\left(\sum_{j=1}^i a_i^2+ 2\sum_{1\leq j < k \leq i} a_ja_k\right) \\&=\sum_{1\leq j \leq i \leq n} a_j^2a_i^2+2\sum_{1\leq j < k \leq i\leq n}a_ja_ka_i^2 \end{aligned}$$ $$\begin{aligned} \sum_{i=1}^{n}\left(a_iS_i\sum_{j=i}^{n}a^2_j\right)&=\sum_{i=1}^{n}\left(a_i\left(\sum_{k=1}^i a_i\right)\left(\sum_{j=i}^{n}a^2_j\right)\right) \\&=\sum_{i=1}^{n}\left(a_i\sum_{1\leq k \leq i\leq j}^{n}a_ka^2_j\right) \\&=\sum_{1\leq k \leq i\leq j\leq n}^{n}a_ka_ia^2_j \\&=\sum_{1\leq i\leq j\leq n}^{n}a_i^2a^2_j + \sum_{1\leq k < i\leq j\leq n}^{n}a_ka_ia^2_j \end{aligned}$$ $$\begin{aligned} \sum\limits_{i=1}^{n}\left(a_iS_i\sum\limits_{j=i}^{n}a^2_j\right)&\le \sum\limits_{i=1}^{n}\left(a_iS_i\right)^2 \\\sum_{1\leq i\leq j\leq n}^{n}a_i^2a^2_j + \sum_{1\leq k < i\leq j\leq n}^{n}a_ka_ia^2_j &\leq \sum_{1\leq j \leq i \leq n} a_j^2a_i^2+2\sum_{1\leq j < k \leq i\leq n}a_ja_ka_i^2 \\0&\leq \sum_{1\leq j < k \leq i\leq n}a_ja_ka_i^2 \end{aligned}$$ Since $a_i$ is non-negative, this inequality is always true, with equality when $a_i=0$ Very nice. Thank YanYau.
Bayern_qing007
19.02.2018 10:16
let \[{b_i} = {a_i}{S_i},{c_i} = \sum\limits_{j = i}^n {a_j^2} ,i = 1,2,...n\]so we just need to prove \[\sum\limits_{i = 1}^n {{b_i}{c_i}} \le \sum\limits_{i = 1}^n {b_i^2} \]notice that for i=1,2,...,n,we have \[\begin{array}{l} {B_i} = {b_1} + {b_2} + ... + {b_i}\\ = {a_1}{S_1} + {a_2}{S_2} + ... + {a_i}{S_i}\\ \le ({a_1} + {a_2} + ...{\rm{ + }}{a_n}){S_i}\\ = S_i^2 \end{array}\]abel LHS, we find that \[\begin{array}{l} \sum\limits_{i = 1}^n {{b_i}{c_i} = \sum\limits_{i = 1}^{n - 1} {{B_i}({c_i} - {c_{i - 1}}) + {B_n}{c_n}} } \\ \le \sum\limits_{i = 1}^{n - 1} {{a_i}S_i^2 + {B_n}{c_n}} \\ \le \sum\limits_{i = 1}^n {{a_i}} S_i^2 = \sum\limits_{i = 1}^n {b_i^2} \end{array}\]
Tyx2019
28.07.2023 15:32
We prove that the coefficients of all terms in the expansion of the LHS is $\leq$ the coefficients of the terms in the expansion of the RHS. Notice all terms in the expansion of the LHS are in the form $a_ia_ja_k^2$, where $1\leq i\leq j \leq k \leq n$(Furthermore all such terms are in the LHS too).
For a fixed $k$, there are $\binom{k}{2}+k$ possible pairs of $i,j$. Hence the total number of such terms are $\sum_{k=1}^n \frac{k(k+1)}{2}$. Furthermore, the total number of terms in the LHS is $\sum_{k=1}^n k(n-k+1)$. Note the coefficients of terms in the LHS are all positive, and hence if $\sum_{k=1}^n k(n-k+1)=\sum_{k=1}^n \frac{k(k+1)}{2}$, all coefficients in the LHS are 1. By some simple algebra, it is clear that this is true, and all coefficients of $a_ia_ja_k^2$ are 1.
Now I will show that all $a_ia_ja_k^2$ "appear" in the RHS. Consider $(a_kS_k)^2$, which is equal to $a_k^2 S_k^2$. Notice $S_k^2$ "contains" $a_ia_j$, where $1\leq i\leq j\leq k$, and hence we are essentially done.