Articles of inequality

Sum of real powers: $\sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$

Let $\{x_i\}_{i=1}^{N}$ be positive real numbers and $\beta \in \mathbb{R}$. Can we say that: $$ \sum_{i=1}^{N}{x_i^{\beta}} \leq \left(\sum_{i=1}^{N}{x_i}\right)^{\beta}$$ I know that this holds if $\beta \in \mathbb{N}$. Does the above inequality have a name in case it’s true?

$C$ such that $\sum_{k\in \mathbb{Z}^n} k_i^2k_j^2|a_{ij}|^2 \leq C \sum_{k\in \mathbb{Z}^n} \|k\|^4|a_{ij}|^2$

More generally, can we find $C_n>0$ such that $$\sum_{k\in \mathbb{Z}^n} k_i^2k_j^2|a_{ij}|^2 \leq C \sum_{k\in \mathbb{Z}^n} \|k\|^2|a_{ij}|^4$$ for all $\{a_k\}_{k\in \mathbb{Z}^n} \in \ell^2(\mathbb{Z}^n)$ where the above sums converge? My ideas so far: $ij \leq i^2 + j^2$ shows that $C=1$ works. Is 1 the sharpest possible bound? (This comes from showing that for $u,f$ periodic and […]

there exist some real $a >0$ such that $\tan{a} = a$

How can i prove that there exist some real $a >0$ such that $\tan{a} = a$ ? I tried compute $$\lim_{x\to\frac{\pi}{2}^{+}}\tan x=\lim_{x\to\frac{\pi}{2}^{+}}\frac{\sin x}{\cos x}$$ We have the situation ” $\frac{1}{0}$ ” which leads us ” $\infty$ ” $$\lim_{x\to\frac{\pi}{2}^{-}}\tan x=\lim_{x\to\frac{\pi}{2}^{-}}\frac{\sin x}{\cos x}$$ We have the situation ” – $\frac{1}{0}$” which tells us ” $- \infty$” This […]

Show that, for all $n > 1: \log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n – 1}$

I’m learning calculus, specifically limit of sequences and derivatives, and need help with the following exercise: Show that for every $n > 1$, $$\log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n – 1} \quad \quad (1)$$ Important: this exercise is the continuation of a previous problem […]

Proving $\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$

This question already has an answer here: Inequality $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ 1 answer

If $a+b+c=3$ show $a^2+b^2+c^2 \leq (27-15\sqrt{3})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

I have become interested in constrained relations among simple cyclic sums involving three positive variables. By simple, I mean so simple that they are also fully symmetric. The “building blocks” of the constraints and relations I have been looking at are: $$ \sum_{\mbox{cyc}} 1 \equiv 3 \\ \sum_{\mbox{cyc}} a \\ \sum_{\mbox{cyc}} ab \\ \sum_{\mbox{cyc}} a^2 […]

Inequalities of expressions completely symmetric in their variables

We often encounter inequalities of symmetric expressions, i.e. the expression doesn’t change if the variables in it are interchanged, with the prior knowledge of a certain relation between those variables. In all such cases that I have encountered thus far, we can find the extremum of the expression by letting the variables equal. Here are […]

If $f'$ is increasing and $f(0)=0$, then $f(x)/x$ is increasing

This question already has an answer here: Show that if $f'$ is strictly increasing, then $\frac{f(x)}{x}$ is increasing over $(0,\infty)$ 2 answers

How prove this inequality $3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\ge 0$

Let $a,b\in[-1,1]$, then prove or disprove: $$f(a,b)=3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\ge 0$$ My try: Since \begin{align*} f(a,b)&=3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\\ &=3ab(a^2+b^2)+18ab(a+b)+12(a^3+b^3)+40(a^2+b^2)+64ab \end{align*} if $ab\ge 0$ then \begin{align*}&f(a,b)\ge 9ab(a^2+b^2)+18ab(a+b)+12(a^3+b^3)+40(a^2+b^2)+52ab\\ &=9ab(a^2+b^2)+18ab(a+b)+12(a+b)(a^2+b^2-ab)+40(a^2+b^2)+52ab\\ &=9ab(a^2+b^2)+6(a+b)(2a^2+2b^2+ab)+40(a^2+b^2)+52ab\\ &\ge0 \end{align*} but for the other case, $ab\le 0$, I can’t proceed. This problem is a follow up from this.

absolute minimum of function

Let $p$ and $q$ be positive numbers satisfying $\dfrac1p+\dfrac1q= 1$; and let $f ∶ [0,+∞) \to \mathbb{R}$ be the function $f(x) =\dfrac1p x^p− x +\dfrac1q$ Show that $f$ has an absolute minimum at $x = 1$ and hence deduce the inequality $ab \le \dfrac1p a^p + \dfrac1q b^q$ for any positive $a$ and $b$. Hint. […]