I am trying to show that $|a+b|\geq||a|-|b||$.  I considered doing an exhaustive process, that is, $a+b \geq 0$, $a+b \leq 0$, and $a+b=0$, and then for each of those considering $|a|-|b| \geq 0$, $|a|-|b| \leq 0$, and $|a|-|b|=0$ (9 combinations)  After doing the first of these ($a+b \geq 0$, $|a|-|b| \geq 0$), I came to the following: $$a+b \geq |a|-|b|$$ and attempting to show this was true.  However, I am unable to show that this is the case.  Additionally, going through such an exhaustive process simply strikes me as wrong or unnecessarily cumbersome.  Any suggestions are appreciated.
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I've seen the full proof of the Triangle Inequality  \begin{equation*} |x+y|\le|x|+|y|.  \end{equation*} However, I haven't seen the proof of the reverse triangle inequality:  \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please prove this using only the Triangle Inequality above?  Thank you very much.
115883
As the title says,  am searching for a proof of      If $A,B \in \mathbb{R}^{n\times n}$ and   $AB=0$ then $\mathrm{rank}(A)+\mathrm{rank}(B) \leq n$   I am doing this as preparation for an upcoming exam and can't figure a way to start. Please just post small hints as answers. I will try to go from there.  Thank you   ftiaronsem
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so as in the title, I have the following theorem to prove.   Theorem  Show that for all $a$, $b\in \mathbb R$, that the following inequality holds, $\begin{equation} ab \leq \frac{1}{2}(a^2 + b^2) \end{equation}$.  My attempt  So if $a=b=0$, then the inequality is trivially $0 \leq 0$ and we are ok.  If $a &gt; 0$ and $b &lt; 0$, then the product on the right is less than zero, while the one on the right is greater than $0$, so the inequality holds.  The problem that I am having is that if the numbers are the same sign, we have that both products will be positive and now we have to the show the inequality is true in this case.   I tried making the following re-arrangement as a starting point $\begin{equation} \\ 2ab \leq (a^2 + b^2) \end{equation}$  But I am not really sure where to go from here. I've been staring at it for a while and I just can't seem to get anything out of it. I noticed that it looks rather similar to the law of cosines for theta equal to 90 degrees and $c = 0$, though that would be wierd indeed. Anyway that hasn't been talked about in the book yet so I don't think it matters.   We have been introduced to the binomial theorem, difference of powers and geometric sum theorems, but I couldn't find any enlightenment from these.  I don't really want an answer if you can help it, but a hint in the right direction would be appreciated if you could offer one. Thanks.
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WTS:  ${n+1\choose k+1}={k\choose k}+{k+1\choose k}+...+{n-1\choose k}+{n\choose k}.$  Algebraically, it is possible, although tedious to show. Is there any combinatorial approach to show this fact?
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Is there a clever way to show that  $$|x| - |y| \leq |x-y|$$  I believe I can think of a way to solve this using cases, but the book I'm working out of said that "A very short proof is possible if you write things in the right way" which has motivated me to figure out this quick way, but I've failed thus far and any help would be appreciated
123884
Given $a_n, b_n$ such that $\exists N   \forall n&gt;N  a_n, b_n \geq 0$ I want to show that if $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge then $\sum_{n=0}^{\infty}a_n b_n$ also converges.  I tried to show it in the following way, if $\sum_{n=0}^{\infty}b_n$ converges, then $\lim_{n\rightarrow\infty}b_n=0$, so there exists $n_0$ such that $b_{n_0}\geq b_n$ for all $n\gt n_0$, therefore $a_n b_{n_0}\geq a_n b_n$, so that $\sum_{n=0}^{\infty}a_n b_n\leq \sum_{n=0}^{n_0}a_n b_n+b_{n_0}\sum_{n=n_0+1}^\infty a_n$, and as my series is bounded by sum of finitely many terms and tail of convergent series multiplied by a constant it's also convergent. Is my proof correct, or are there any flaws in my reasoning?
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The question reads     $$U_n = (1+x)^n - 1 - nx$$   Show that $U_2 \geq 0$   Hence or otherwise show that $(1+x)^n \geq 1 + nx$ for all $x \gt -1$.   Obviously the $U_2 \geq 0$ is very easy, I can do that without any trouble but I cannot see how it links to the 2nd part of the question.
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I am given that $${n \choose a}={n \choose b}$$ I need to show that $n=a+b$ or $a=b$.  Now I am left with $a!(n-a)!=b!(n-b)!$.  I don't know what to do next. I checked there on google to solve the above, it says:- "On comparing, we get the required result." But it makes no sense to me.
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The constraints are   $0 \le a_1,a_2....a_n,b_1,b_2....b_n \le 1$. $|a_i-b_i|&lt; \delta$ for all $1 \le i \le n $   How do I go about proving the following $$|\prod_{i=1}^n a_i-\prod_{i=1}^n b_i| &lt; n\delta$$  I tried reducing it to two terms where one term is like $(a-\delta)^n$ so that I can get the $n\delta$ term from binomial,but I am stuck. I would really appreciate just hints.
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Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$   example1: $|x-1|+|x-2|\geq 1$  my solution：(substitution)  $x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$  square,  $t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$  so proved.  question1 : Is my proof right? Alternatives?  one reference answer:   $1-|x-1|\leq |1-(x-1)|=|1-x+1|=|x-2|$  question2 : prove:  $|x-1|+|x-2|+|x-3|\geq 2$  So I guess:( I think there is a name about this, what's that? wiki item?)  $|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$  How to prove this? This is  question3. I doubt whether the two methods I used above may suit for this general case.  Of course, welcome any interesting answers and good comments.
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Could anyone explain how this statement is true? You may notice that this statement is part of the process of adding two independent binomial r.v.'s.  $$ \ \sum_{x=0}^\infty{n \choose x}{m \choose w-x} ={n+m \choose w}\ $$
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Let $a$ and $b$ be positive numbers, and $n \in \mathbb{N}$. Prove that (using Rearrangement Inequality) $$(n+1)\left(a^{n+1}+b^{n+1}\right) \geq (a+b)\left(a^n+a^{n-1}b+\ldots+b^{n}\right). $$  Thanks :)
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Let $a$, $b$, and $c$ be integers, where a $\ne$ 0. Then $$ $$ (i) if $a$ | $b$ and $a$ | $c$, then $a$ | ($b+c$) $$ $$ (ii) if $a$ | $b$ then $a$|$bc$ for all integers $c$; $$ $$ (iii) if $a$ |$b$ and $b$|$c$, then $a$|$c$.  Prove that if $a$|$b$ and $b$|$c$ then $a$|$c$ using a column proof that has steps in the first column and the reason for the step in the second column.  My book is really vague. I'm not really sure what to do..  First I was thinking something like this: $a\mid b\Rightarrow b=as$ and  $b\mid c\Rightarrow c=bt$ and $a\mid c\Rightarrow c = au$ $$ $$ $b + c + c = b + 2c$  $$ $$ $=as + 2(bt+au)$  Yea then I just got lost.
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Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$  Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.  My idea:  I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.  Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$ I have to show that $$y^2-2xz\leq 3$$ I tried to prove it with the sign of trinom, but it doesn't work.
80543
Let be $a, b$ in $(0,1)$ such that $a+b&gt;1$. I need to prove that:  $$2^a+3^b&lt;3a+4b$$  I'm looking for an elementary proof that doesn't resort to the calculus tools.
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If I know for two numbers a and b that    $${a &lt; b }$$  Then is it correct to say that    $$ a \leq b $$  I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information.
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Can you help me to prove that $$|x-y|\leq |x|+|y|$$ I get a proff of this equality, but it's very short and I don't know if it's correct.