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## Hot answers tagged inequalities

34

Integrate by parts: \begin{align} \int_x^{x+1}\sin(e^t)dt & =\int_x^{x+1}e^{-t}d(-\cos(e^t)) \\ & =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-t}\cos e^{t}dt\\ & =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-\int_x^{x+1}e^{-2t}d\sin e^{t}\\ & =e^{-x}\cos e^x-e^{-x-1}\cos e^{x+1}-e^{-2(x+1)}\sin e^{x+1}\\ & \hphantom{={}}+e^{-2x}\sin e^x+2\...

26

The class of concentration of measure inequalities is a fundamental tool in modern probability (and any field that uses probability, e.g., random matrix theory, theoretical computer science, statistics, high-dimensional geometry, combinatorics, etc.). As explained in this blog post of Scott Aaronson, these are basic ways in which one "upper bounds the ...

18

As suggested by Joe Silverman, there are counterexamples in my paper Rational points near curves and small nonzero $|x^3-y^2|$ via lattice reduction, Lecture Notes in Computer Science 1838 (proceedings of ANTS-4, 2000; W.Bosma, ed.), 33-63. arXiv: math.NT/0005139 (https://arxiv.org/abs/math/0005139) The best one there, which I think still holds the ...

18

For every $n\in\{1,\dotsc,N\}$, we have $$2\sqrt{\sum_{i\leq n} a_i}-2\sqrt{\sum_{i\leq n-1} a_i}=\frac{2a_n}{\sqrt{\sum_{i\leq n} a_i}+\sqrt{\sum_{i\leq n-1} a_i}}>\frac{a_n}{\sqrt{\sum_{i\leq n} a_i}}.$$ Summing these up, we obtain the inequality with $C=2$. It is also straightforward to see that for $C<2$ the inequality fails, hence $C=2$ is the ...

16

As noted in the comments, this is a version of a conjecture originally made by Marshall Hall many years ago. The original conjecture was that there is a constant $k$ so that if $m^2\ne n^3$, then $$|m^2-n^3| > k \sqrt{|n|}.$$ As noted on Elkies' webpage (http://people.math.harvard.edu/~elkies/hall.html), this is widely believed to be false. In any case,...

16

Gaussian Jensen's inequality: Let $\boldsymbol{X}=(X_{1}, \ldots, X_{n})\sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol\Sigma)$ be a gaussian vector. The inequality $$\mathbb{E} B(f_{1}(X_{1}), \ldots, f_{n}(X_{n})) \leq B(\mathbb{E}f_{1}(X_{1}), \ldots, \mathbb{E}f_{n}(X_{n}))$$ holds for all real valued (test functions) $f_{1}, \ldots, f_{n}$ if and only ...

14

Rewrite your inequality as $$lhs:=\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}\le C\sqrt{s_N},$$ where $s_n:=\sum_{i=1}^n a_i$. Note that $$\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}$$ is a lower Riemann sum for the integral $$\int_0^{s_N}\frac{ds}{\sqrt s}=2\sqrt{s_N}.$$ So, $$lhs\le2\sqrt{s_N},$$ as desired. In the above proof, it was tacitly assumed ...

14

Yes, because the OP stated that the ground field is $\mathbb{R}$, one can simply take the octic polynomial $$Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i < j\le n} \bigl((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2\bigr)^2,$$ which will do the trick.

13

$d_2$ is indeed a metric. Abbreviating $\gcd(m,n)$ to $(m,n)$, we need to show that \begin{align*} 1-\frac{2(a,c)}{a+c} &\le 1-\frac{2(a,b)}{a+b} + 1-\frac{2(b,c)}{b+c} \end{align*} or equivalently \begin{align*} \frac{2(a,b)}{a+b} + \frac{2(b,c)}{b+c} &\le 1 + \frac{2(a,c)}{a+c}. \end{align*} Furthermore, we may assume that $\gcd(a,b,c)=1$, since we ...

13

Without loss of generality, $u_1,\dots,u_n$ can be taken to be the standard basis of ${\bf R}^n$ (so $e_1,\dots,e_r$ is just some arbitrary orthonormal system). We can view $\Lambda$ as a real symmetric $n \times n$ matrix of Frobenius norm $1$ and rank at most $r$. (Conversely, every such matrix has a representation of the desired form for some $e_1,\dots,... 13 Your conjecture is indeed correct. The proof is based on the following simple yet powerful trick I learned many years ago:$|z| = \sup_{|v| = 1} \Re (zv)$. Therefore it is enough to only bound from above$\Re \left(\int_0^1 vf(x)e^{-2\pi ix}dx\right)$for all$v\in \mathbb{T}$. And now it is clear by Cauchy-Schwarz that$f$should be proportional to$\max(0, ...

12

This is a supplement (correction) to Peter Mueller's nice solution. As he observed, it suffices to show that, for any fixed $n\geq 1$ and $y\in[0,1]$, the function $$z\mapsto\left(\frac{y^{n+1}+z}{y^{n-1}+z}\right)^n+\left(\frac{zy^{n+1}+1}{zy^{n-1}+1}\right)^n,\qquad z\in(0,1),$$ is increasing. (Indeed, $y:=x^{1/2}$ and $z:=x^{(n-1)/2}$ yields the LHS of ...

11

Here is a method that will allow one to find the exact upper and lower bounds on $g(z)$ over $z>0$ with any degree of accuracy. Take any real $z>0$. Since \begin{equation*} \frac1y=\int_0^\infty dt\,e^{-y t} \end{equation*} for any real $y>0$, we have \begin{align*} \frac{g(z)}z &=\int_z^{e z} dy\, \frac{\sin y}y \\ &=\int_0^\infty ...

10

You can prove it using the Golden-Thompson inequality $Tr (e^{A+B}) \leqslant Tr(e^{A} e^{B})$ and then applying the Cauchy-Schwarz inequality.

10

The inequality is false for $a=1799, b=105, c=1024, d=4$. This counterexample was found and verified with Mathematica, as follows:

9

Take $n=3k$, $2k$ variables equal to $3$ and $k$ variables equal to $-5$ for large $k$. Then $\sum a_i=k>0$, and $\sum_{i<j<k} a_ia_ja_k=\frac16 (\sum a_i)^3+O(k^2)=\frac{k^3}6+O(k^2)>0$ for large $k$. But $\sum a_i^3<0$.

9

An LMFDB search for curves with many integer points turns up the curve 20888a1: $y^2 = x^3 - 52 x + 100$ which has $52$ integral points, a bit more than your conjectured bound of about $47.052$ using $(M_x, M_y) = (12214, 1349854)$. It does seem to be true that curves with many integer points tend to have a few large ones, but I don't know of any ...

8


8

Let $k:=\pi(n)$, so that $p_k\le n<p_{k+1}$, where $p_k$ is the $k$th prime. By the last displayed formula in this section of the Wikipedia article, \begin{equation*} -1+\ln(k\ln k)<\frac{p_k}k<\ln(k\ln k) \end{equation*} if $k\ge6$, whence \begin{equation*} n>-k+k\ln(k\ln k),\quad n/2<m_k:=\frac{k+1}2\,\ln((k+1)\ln(k+1)). \end{equation*} ...

8

$$\|\sqrt{A+B}Cx\|^2=(\sqrt{A+B}Cx,\sqrt{A+B}Cx)=\\ ((A+B)Cx,Cx)=(ACx,Cx)+(BCx,Cx)=\|\sqrt{A}Cx\|^2+\|\sqrt{B}Cx\|^2,$$ taking the supremum over unit vectors $x$ we get $$\|\sqrt{A+B}C\|^2\leqslant \|\sqrt{A}C\|^2+\|\sqrt{B}C\|^2\leqslant (\|\sqrt{A}C\|+\|\sqrt{B}C\|)^2.$$

7

A counterexample: $n=4$, $$(x_1,\dots,x_4)=\left(-\frac{1}{4},-\frac{3}{8},-\frac{3}{8},\frac{\sqrt{1970156929}-2048}{45375}\right).$$ So, by what you noted, your conjecture fails to hold for any $n\ge4$. The validity gap is indeed between $n=3$ and $n=4$. Indeed, for $n=3$ Mathematica confirms your conjecture: I think this can also be easily checked ...

7

Let me rename $p$ to $z$, because $p$ usually stands for prime numbers in the subject. I will assume that $z\geq 2$, but I will not assume that $z$ is an integer. The Dirichlet coefficients of $\zeta(s)^z$ form a generalized divisor function: $$\zeta(s)^\nu=\sum_{n=1}^\infty\frac{\tau_z(n)}{n^s},\qquad \Re(s)>1.$$ The generalized divisor function is ...

7


7

There is no real $C>0$ such that $$\sqrt{T^2+S^2} \le C(T+S) \tag{1}$$ holds for all positive definite (self-adjoint) operators on a Hilbert space of any dimension $\ge2$. Indeed, take any any real $C>0$ and identify $T$ and $S$ with the $2\times2$ matrices $$T:=\left( \begin{array}{cc} 1 & 0 \\ 0 & s^5 \\ \end{array} \right),\quad S:=\... 7 One can transfer the continuous L^p theory to this discrete setting without difficulty. Let's normalise \sum_k |a_k|^p = \sum_n |b_n|^q = 1. Consider the two quantities$$ X_1 := \sum_{k \neq n} \frac{a_k \overline{b_n}}{\lambda_k - \lambda_n} X_2 := \sum_{k, n} p.v. \int_{{\bf R}^2} \varphi(s) \varphi(t) \frac{a_k \overline{b_n}}{(\lambda_k+s) -...

7

The classical operator generalization of the scalar inequality $|\sqrt{a}-\sqrt{b}|^2 \leq |a-b|$ is the Powers-Størmer inequality, which involves two different norms : the trace norm $\|X\|_1 = \operatorname{Tr}|X|$ and the Froebenius norm $\|X\|_2 = (\operatorname{Tr}(X^* X))^{\frac 1 2}$, where $|X| = (X^* X)^{\frac 1 2}$ is the usual absolute value of ...

7

Write $$G=\left( \begin{array}{ccc} a & b & c \\ b & d & e \\ c & e & f \\ \end{array} \right)$$ and, without loss of generality, $$U=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & u & 0 \\ 0 & 0 & v \\ \end{array} \right),$$ where $u,v>0$. Then $$\text{Tr}(GUGU^{-1})-\text{Tr}(G^2) =\frac{b^2 (u-1)^2 v+... 6 I am afraid it is false. Take F=0, A=C, E=2A, \mu=1/2. Then we are given \cos B+\cos D=2\cos A and should prove \cos B/2+\cos D/2\geqslant 2\cos A/2. But if \cos B=x, then \cos B/2=\sqrt{(1+x)/2}, this function is concave, thus inverse inequality \cos B/2+\cos D/2\leqslant 2\cos A/2 holds. 6 Answer to Version 1 of the question: No, let b_1,b_2,a_1,a_2 be 2,6,3,4. Then$$ (1-1/2)(1-1/6)=5/12 \not\ge 1/2=(1-1/3)(1-1/4). $$6 This is the isoperimetric inequality in disguise. Define the Riemannian metric with length element e^{u(z)}|dz|. The curvature of this metric is$$-e^{-2u}\Delta u=0, so the metric is flat. Then your inequality says that the area of the disk is at most $1/4\pi$ times the square of the length of the boundary, the usual isoperimetric inequality for the ...

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