3
votes
1answer
31 views

Is the norm of the average $\le$ the norm of the max?

Given $\pmb X \in \mathcal{R}^p$, denote the elements of $\pmb X$ as $\pmb x_i$ for $i= 1, \dots, n$. Denote the $t(\pmb X)$ as the average of $\pmb X$ \begin{equation} \pmb t(\pmb X) = \frac 1 n ...
1
vote
2answers
60 views

How to establish this inequality: $(1-a)(1-b)(1-c) \geq 8abc$ for $a+b+c=1$?

Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$ My effort: Since $a+b+c =1$, we can write $$ ...
2
votes
1answer
44 views

What is the most elementary proof of these inequalities?

Let $p$ be a non-zero integer, and let $x_1$, $\ldots$, $x_n$ be $n$ positive real numbers. Then we define the $p$-th power mean $M_p$ of these numbers as $$ M_p \colon= (\frac{x_1^p + \ldots + ...
2
votes
1answer
48 views

Maximum and minimum of weighted sum

For $w_i\ge 0$ and some constants $\alpha_i , i=1,...,n$, what is the maximum and minimum of $\sum_{i=1}^{n}\alpha_i w_i$ subjected to $\sum_{i=1}^{n}w_i=1$? Intuitively, I put all weight on the ...
0
votes
1answer
29 views

Properties of weighted average ratio of weights

Suppose we have two sets of weights $w_i$ and $u_i$; i.e., $\sum w_i =\sum u_i = 1$, and $0 \le w_i \le 1$, $0 \le u_i \le 1$. Consider the weighted average ratio R of these weights: $R \equiv \sum ...
3
votes
1answer
62 views

Is it always true that “max $\ge$ average + sigma”?

Assume that $i$ from $1,\ldots,N$, $x_i \ge 0$ and: $$\mathrm{avg} = \frac{\sum_i x_i}{N}$$ $$\sigma = \sqrt\frac{\sum_i{(x_i-\mathrm{avg})^2}}{N}$$ Is that true that: $$\max_i x_i \ge ...
1
vote
2answers
171 views

Average limit superior [duplicate]

Let $\mathcal{l}_\mathbb{R}^\infty$ be the space of bounded sequences in $\mathbb{R}$. We define a map $p: \mathcal{l}_\mathbb{R}^\infty\to\mathbb{R}$ by $$p(\underline x)=\limsup_{n\to\infty} ...
3
votes
4answers
177 views

If $\sum\limits_{k=1}^n y_k\geq n$ and $\sum\limits_{k=1}^n \frac{1}{y_k}\geq n$, then $\prod\limits_{k=1}^n y_k\geq 1$?

Let $y_1,\ldots y_n$ be positive real numbers satisfying $y_1+\cdots+y_n\geq n$ and $\displaystyle{\frac{1}{y_1}+\cdots+\frac{1}{y_n}\geq n}$. Is it true that $y_1y_2\cdots y_n\geq 1$?
1
vote
1answer
203 views

Bound on deviation between arithmetic and harmonic mean?

It is well known that, if HM denotes the harmonic mean and AM the arithmetic mean, we have $$ AM(x) \ge HM(x) $$ Now I am dealing with the expression $$ \frac{1}{HM(x)} - \frac{1}{AM(x)} $$ A ...