Not sure how to proceed. I'm trying to prove that the following inequality is true. I know that $t_2 = 6$ and $t_3=17$ from the problem statement. The base case is obvious.   $t_{r+1} \leq (r+1) (t_r - 1) + 2$  ($t_r$ is supposed to be the  $R(3,3,3,\ldots,3)$ with $r$ $3$s).
120920
I want to prove that:  $$R(\underbrace{3,3,...,3,3}_{k+1}) \le (k+1)(R(\underbrace{3,3,...3}_k)-1)+2$$  where R is a Ramsey number. In the LHS, there are $k+1$ $3$'s, and in the RHS, there are $k$ $3's$. I really have no clue how to start this proof. Any help is appreciated!
80304
We have been using this result without proof in my class, but I don't know how to prove it. Could someone point me in the right direction?  $$|1-e^{i\theta}|\le|\theta|$$  I believe this is true for all $\theta\in\mathbb{R}$. It is easy to show that the left side is bounded by 2 (triangle inequality), but I'm stuck otherwise.
91951
Any hints on how I can prove the inequality:     $$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$   Thank you.
120815
So, here is the question:  For any position integer $n$, let $T(n)$ be the number 1 if $n&lt;4$ and the number $T(n-1) + T(n-2) + T(n-3)$ if $n \geq 4$.  We have $T(1)=1, T(2)=2, T(3)=3$ $$T(4)=T(3)+T(2)+T(1) = 1+1+1+1 = 3$$  $$T(5) = T(4)+T(3)+T(2) = 3+1+1 = 5$$  Prove that: (Universal n is an element of all positive integers), $T(n)&lt;2^n$  Any suggestions?
15535
Problem: Show that if $x_n\to 5$ then $\dfrac{x_n +1}{\sqrt{x_n -1}} \to 3$.  So I know $|x-5|&lt;\epsilon$ and I need to show that $\left|\dfrac{x_n +1}{\sqrt{x_n -1}} -3\right| &lt; \epsilon$.  I tried applying the triangle inequality (and a bit of algebra) to get $$\left|\frac{x_n +1}{\sqrt{x_n -1}} -3\right| \leq |x_n -5| \left |\frac{1}{\sqrt{x_n -1}}\right| + \left|\frac{6}{\sqrt{x_n-1}} -3\right|.$$  From this I know that $|x_n -5| &lt;\epsilon$ and can show that $\left |\dfrac{1}{\sqrt{x_n -1}}\right|&lt;1$. So my only problem is the second part of the sum. I have no idea what to do with it to get everything to be less than $\epsilon$.
74826
Show that $\max{\{|a|+|b|,|c|+|d|\}} \leq \max{\{|a|,|c|\}}+\max{\{|b|,|d|\}}.$  I wanted to show that $d(p,q)=\max{\{|x_1-x_2|,|y_1-y_2|\}}$ where $p=(x_1,y_1),q=(x_2,y_2)$ is a metric on $\mathbb{R^2}.$ In proving the triangle inequality let $p=(x_1,y_1),q=(x_2,y_2),r=(x_3,y_3).$ Then  \begin{align*} d(p,r)&amp;=\max{\{|x_1-x_3|,|y_1-y_3|\}}\\ &amp;\le \max{\{|x_1-x_2|+|x_2-x_3|,|y_1-y_2|+|y_2-y_3|\}}\\ &amp;\le \max{\{|x_1-x_2|,|y_1-y_2|\}}+\max{\{|x_2-x_3|,|y_2-y_3|\}}=d(p,q)+d(q,r) \end{align*}  I don't understand how this inequality is true.
36710
Let $a,b,c$ real numbers. Prove the inequality $|a-b| \leq |a-c| + |c-b|$. Prove that equality holds if and only if $a \leq c \leq b$ or $b \leq c \leq a$.   I've tried starting with just $a \leq c$ and using field properties to reconstruct the inequality, however I haven't been able to make it work. I also tried making the negatives positive and stripping the inequalities and making something happen but again I don't know if that's a proper rule and it didn't seem to get me anywhere.
116601
If $a$ and $b$ are non-negative real numbers then show that $$(3a +\frac{4}{a+1}+\frac{8}{\sqrt{2(1+b^2)}})\cdot(3b +\frac{4}{b+1}+\frac{8}{\sqrt{2(1+a^2)}})\geq81.$$ Inequality is less common. To demonstrate the usual instruments (means inequality, ...) did not help. Does anyone have any idea?
70349
Prove that $|a| \leq \max\{|b|,|c|\}$ if $b\leq a \leq c$.  I showed that $a\leq c$ and therefore $-c\leq a \leq c$ so that $|a|\leq c$ but then I got stuck. Is this the right approach?
26939
I am having trouble with this problem from my latest homework.  Prove the arithmetic-geometric mean inequality. That is, for two positive real numbers $x,y$, we have $$ \sqrt{xy}≤ \frac{x+y}{2} .$$ Furthermore, equality occurs if and only if $x = y$.  Any and all help would be appreciated.
51490
I stumbled on the following inequality: For all $n\geq 1,$ $$2\sqrt{n+1}-2\sqrt{n}&lt;\frac{1}{\sqrt{n}}&lt;2\sqrt{n}-2\sqrt{n-1}.$$ However I cannot find the proof of this anywhere. Any ideas how to proceed?  Edit: I posted a follow-up question about generalizations of this inequality here:
39982
I have an inequality problem which is as follow:  If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.  I am not so good in inequalities. So, please give me some hints so that I can proceed.  Thanks.
91390
how do you prove this inequality? $|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}$ At first glance I thought the triangle inequality would be useful but it's not form what I see. If I raise everything to the power of n I get the following: Possibility 1 (n uneven) : $|x + ... -y| \le |x-y|$ I guess I would have to prove that ... $\le$ 0 but I don't know how to do that. Possibility 2 (n even) : $|x + ... +y| \le |x-y|$ Same as above but now I have to prove that ... $\le$ -y. Am I even on the right track here? Any help would be appreciated. Thanks in advance.
118439
Given $a, b \in \mathbb{R}$, prove that   $$\frac{|a + b|}{1 + |a + b|} \le \frac{|a|}{1 + |a|} + \frac{|b|}{1+|b|}$$  When does equality hold?  The only useful thing I could get (using the triangle inequality) is:  $$\begin{align}1 + \left|\left(a + \frac{ab}{2}\right) + \left(b - \frac{ab}{2}\right)\right| &amp;\le 1 + \left|a + \frac{ab}{2}\right| + \left|b - \frac{ab}{2}\right|\\ &amp;\le 1 + |a| + |b| + |ab|\\ &amp;=(1 + |a|)(1 + |b|)\end{align}$$  But it seems to be inapplicable! Any ideas?
82957
Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$, then for any choice of positive integers $a$ and $b$,  $t^3$ never equals to a) $4t+3$  b) $8t+5$  c) $10t+3$  d) $6t+5$  I have recognized that if $t=0$ then all the answers are correct.  But can not proceed in case of $t\neq 0$. Please help to solve it. Thanks in advance.
40300
Let $T : [0, 1]^2 → [0, 1]$. Consider the following properties:  $T_1 : T(x, 1) = x$  $T_2 : T(x, y) = T (y, x) $  $T_3 : T(x, T (y, z)) = T(T (x, y), z) $  $T_4 $: If  $x ≤ u  , y ≤ v \Rightarrow T(x, y) ≤ T(u, v) $  Function $T: [0, 1]^2 → [0, 1]$ that satisfies $T_1 −T_4$ is T-norm . For $T(x) = \begin{cases}min(x,y),&amp;max(x,y)=1,\\0,&amp;o.w.\end{cases} $ conditions 1, 2 and 4 obviously.  Can you help me prove condition 3.
13096
Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) &gt; (a+b+c)(a^2+b^2+c^2)$.  I know the proof by subtracting LHS by RHS and then doing some arrangement.  But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) &gt;(a+b+c)(a^2+b^2+c^2)$, or an easy proof?