I have seen things like \lessgtr $\lessgtr$ (less on top, greater on bottom;) being used in paper. Can someone give me an example of how a double inequality is used?
2719
I'm reading Papadimitriou &amp; Steiglitz's Combinatorial Optimization and came across notation I'd never seen before and don't know what it means. The $\LaTeX$ markup for it is \gtrless ($\gtrless$), which took me quite a while to find.  It arises in the formulation of general linear programs in terms of the constraints on the variables:  $$ x_j \geq 0 \;\; j \in N\\ x_j \gtrless 0 \;\; j \in \bar{N} $$  It's not "not equals" because there's places in the text where the authors say $x$ can be zero.
122087
In english based math language it seems that non-increasing $\Longleftrightarrow$ less or equal  (non-strict decreasing) decreasing     $\Longleftrightarrow$ strict less    (    strict decreasing) Is that correct ? If so, how does it make sense ? precision I should note that even very good math teacher are making mistakes about this. Actually I asked this question after watching Boyd's video on convex optimization where even him is confused about this.... So I imagine many many people are, and there must be classes and tests about this absurd and buggy concept, which yields absolutely nothing interesting. So I just wonder if I really am missing something, or if, yes, some people decided to create an abstraction that is leaky (not not increasing $\neq$ increasing ?) verbose (4 words, with special negation logic, instead of using the word 'strict'  and keeping the usual well defined predicate logic rules) absurdity of the concept This notation is absurd for the following reason :  when dealing with element instead of functions we dont apply the same logic : we dont phrase $x &lt; y$ as  &quot;$x$ is less than $y$&quot; nor  &quot;$x\leq y$&quot; as &quot;$x$ is not-more than $y$&quot;. (If we did though, at least it would not be so harmful as not not-more would mean more)  you have to define functions using a not notation, $f$ is non-increasing function $\Longrightarrow$ if $x$ is not-less than $y$, say 0.3 feet and 2.5 inches, then $f(x)$ is not-more than $f(y)$   This also violates a very basic tenet in programming style 101, which is here for a reason : never define or use something with a negation, it is confusing.  To apply composition rules between functions, you better be buckled up with all the not. must be a fluff of cases  More profoundly, this violates a fundamental principle of logic which is that given some ambiguity, you should assume the most general case apply. It is way worse than measuring things with non integral units. This is violating logical rules, and leaving a very basic concept obfuscated.
44288
I know some of it, but I would like to learn other ones. I think this question is useful, as it will be helpful for other beginners. Such as: uvw, AM GM, Cauchy–Schwarz inequality, Jensen inequality. Could you please add something? Maybe books containing it?
21748
I am having issues solving these problems: I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.     Show that for $a,b\in (-\infty,+\infty)$,   the following inequalities apply:   a) $2|ab| \leq a^2 + b^2$  b) $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$
54078
Let $x,y,z&gt;0$ and $x+y+z=1$, then find the least value of  $${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$  I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help me.  I wrote $x$ as $1-(y+z)$ and I took $x+y$ as $a$ and the others as $b$ and $c$. And I am trying.
74314
I want to lower bound \begin{align}  \mathbb{P} \left( -\epsilon \leq  \sum_{i=1}^n X_i \leq 0 \right) &amp;= \mathbb{P} \left( - \frac{\epsilon}{\sqrt{\sum_{i=1}^n \sigma_i^2}} \leq  \frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n \sigma_i^2} \leq 0 \right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \end{align} where $\{X_i\}$ are mean zero, independent random variables with $\mathbb{E}[X_i^2] = \sigma_i^2 \leq H^2$ and $\mathbb{E}[|X_i|^3] = \lambda_i^3 \leq H^3$. From Equation $3$ in , I have that \begin{align} \sup_{x \in \mathbb{R}} |F_n(x) - \Phi(x)| \leq C_0 \left( \sum_{i=1}^n \sigma_i^2 \right)^{-3/2} \sum_{i=1}^n \lambda_i^3 \end{align} Loosening the bound above gives \begin{align} \sup_{x \in \mathbb{R}} |F_n(x) - \Phi(x)| \leq \frac{C_0}{\sqrt{n}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2) \end{align} Now I can write $(1)$ as \begin{align} F_n(0) - F_n \left(-  \frac{\epsilon}{\sqrt{\sum_{i=1}^n \sigma_i^2}} \right) \end{align} which by $(2)$ can be lower bounded as \begin{align} \Phi(0) - \frac{2C_0}{\sqrt{n}} - \Phi\left ( -  \frac{\epsilon}{\sqrt{\sum_{i=1}^n \sigma_i^2}} \right ) &amp;\geq \frac{1}{2} - \frac{2C_0}{\sqrt{n}} - \Phi\left ( -  \frac{\epsilon}{H \sqrt{n}} \right ) \end{align} Can the last expression be improved to something like $K/\sqrt{n}$ for some constant $K$ for large $n$? I feel like I need some approximations to normal CDF $\Phi(x)$ for small $x$.
102104
I am writing some proofs using LaTeX, and following the standard procedure in the ConTeXt, I need to write something as follows:    However, I have no idea how to create the "double bracket", like the one includes x := y + z...  Could anyone give me some help on this?
49788
If $a, b \in \mathbb{C}$, then we have the standard triangle inequality for the difference:  $$||a| - |b|| \le |a - b|.$$  I am wondering if this inequality generalizes to exponents greater that one.     My question is, for $1 &lt; p &lt; \infty$, does there exists a constant $C_p$ such that $||a|^p - |b|^p| \le C_p|a - b|^p$ for all $a, b \in \mathbb{C}$?   I am aware of the "standard" triangle inequality for $p &gt; 1$:  $$|a +b|^p \le 2^{p}(|a|^p + |b|^p),$$  and that $2^p$ may not be sharpest constant possible. If I try to use this estimate to prove, say, that $|a|^p - |b|^p \le C_p |a - b|^p$, I get stuck with an extra term that I'm not sure what to do with.  $$|a|^p - |b|^p \le 2^p|a -b|^p + (2^p -1)|b|^p.$$  A resolution on this matter is greatly appreciated.
22270
Why is the following inequality from Chapter 1 of Littlewood's  true?      Suppose $a_n&gt;0$ for all $n$. Then $$\limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n\geq e.$$
86257
Proof that for $a,b,c\in \mathbb{R}$, hold that $\left |\sqrt {a^2+b^2}-\sqrt {a^2+c^2}  \right |\leq \left |b-c  \right |$. Give me a clue, I don't have any idea to work with that inequality.
98279
I have derived the inequality if $k&gt;1$,  ${n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k&lt;{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k$  But, my problem is how to use this inequality to prove that if $n\geq1$, $(1+{1\over n})^n&lt;(1+{1\over n+1})^{n+1}$
54747
I need to lower bound the tail probability of a non-negative random variable. I have a lower bound on its expected value. I am aware of a reverse markov's inequality that does the job when the random variable is bounded above. Unfortunately that is not my case.   Is there any other inequality that may be useful to me in this regard?  thanks  NR
36735
Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time.
88250
I proved this inequality in the following way:   Lemma: $r \in \Bbb N, r \geq 3$. We have $r^r \gt (r+1)^{r-1}$.  Proof: We apply the AM-GM inequality to the $r$ positive integers where there are $r-1$ $(r+1)$'s and one $1$. We obtain $$\frac {1+(r-1)(r+1)}{r} \gt  ((r+1)^{r-1})^{\frac {1}{r}}$$ wherefrom we get (since the $r^{{\rm th}}$ power function is increasing), $r^{r} \gt (r+1)^{r-1}$.   Now, I used mathematical induction to prove the statement. We have from the lemma, $(k!)^{2} \gt k^k \implies (k!)^{2} \gt (k+1)^{k-1}$ and multiplying this inequality by $(k+1)^{2}$, we have $((k+1)!)^{2} \gt (k+1)^{k+1}$ and obviously $(3!)^{2} \gt 3^3$.  Is there any direct proof of the statement that does not use induction and calculus?
59750
I want some advice about how to format it better if I have to write a long text below a $\sum$ or a $\prod$ sign. Basically I want to explain the restriction over which the sum or the product is being taken.   For example how can I improve the looks of these equations,  $N_i (\rho) = \sum _ {B \in \rho s.t n_{\delta_j}(B) = 0    \forall j \neq i} n_{\delta_i} (B)$  $\sum _{ \rho s.t N_i(\rho) = a_i and M_i(\rho) = b_i}    \prod _{B \in \rho} \langle \prod_{i=1}^{n} \delta_i ^{n_{\delta_i}(B)}    y_i ^{n_{y_i}(B)} \rangle_c$
48332
I'm looking for a reference for a matrix-norm inequality that I used in , which has a few equivalent forms.  I will use notation that applies to complex vector spaces with a sesquilinear inner product, but of course the same applies over real matrices.  The statement is as follows:     Take $A,B \in \Bbb F^{n \times n}$.  Then    $$\vert\operatorname{tr}(A^*B)\vert \leq \sigma_1(A)\sum_{i=1}^n \sigma_i(B) = \|A\| \operatorname{tr}|B|$$   where $\sigma_i$ denotes the $i$th singular value, $|B| = (B^*B)^{1/2}$, and $\|\cdot\|$ denotes the spectral norm (induced Euclidean norm).   I did manage to find some references, but they're overkill, and the texts themselves are not readily accessible to the faint of heart (Bhatia's text is dense and Pedersen's is not about matrices in particular).    A suitable reference would be greatly appreciated.
122038
Let $\{X,\mathcal{A},\mu\}$ be a probability space. Let $A$, $B$, $C$ be sets in $\mathcal{A}$, which satisfy $$\mu (A\cap B)\geq \frac{1}{2}, \quad \mu (A\cap C)\geq \frac{1}{2}, \quad \mu (B\cap C)\geq \frac{1}{2}.$$ Show that $\frac{1}{4}\leq \mu(A\cap B \cap C)$.  I've managed to show in a previous assignment that $$\mu(A\cap B \cap C)\leq \min\{\mu (A\cap B), \mu (A\cap C), \mu (B\cap C)\},$$ but now I'm stuck with the above inequality. I've tried to draw it, but still no luck. Any hints?