I think that for positive natural numbers t and n we have  $$ \left(1+\frac{n}{t}\right)^t\ \le\ e^n\,. $$   Is this true? I have constructed a proof of it (which would probably take some lengthy typesetting here, and further I believe that the inequality could be very standard), but I wish to have an independent confirmation from the folks here. If the answer to the previous question is "yes", is there a proof in the published, reviewed literature (book, article, thesis, etc.) that I can reference? We really don't want to show that we've reinvented the wheel.   EDIT: Since the answers confirmed the inequality, and there are even duplicate questions, I'm really interested in part 2 from now on.
123027
I am having trouble with my homework problem, it says:      Suppose that $n$ is a positive integer and that $x &gt; 0$. Show that   $$\left(1+\frac{x}{n}\right)^n &lt; e^x.$$    I have proved the base case, but I am unsure how to proceed with the induction step. The question continues:      and, if $x &lt; n$, then   $$e^x &lt; \left(1-\frac{x}{n}\right)^{-n}$$   Also, by choosing a suitable $n$, deduce that $2.5 &lt; e &lt; 2.99$.   I would like a hint to help me along if anyone can help. Thanks!
100760
So I've spent a few hours trying to find a counter example to this statement, and then gave up and spent additional few hours trying to prove it and Ive finally became frustrated with this problem. :) I seem to get stuck in the scenario where $f\left(\frac{1}{x}\right)$ &lt; 0, I'm just unable to find an epsilon that will satisfy all the conditions.. plus I'm not even sure if the statement is true or false anymore..
115559
I have been trying to show these two inequalities hold for all positive integers n, but I don't know how to proceed at all... I have tried playing around with them but I didn't find anything helpful. I have also tried induction but I couldn't make it work.
36572
I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:  $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$  I would appreciate your help.
121264
I want to prove that for each real number $x&gt;0$, the statement  $${\left(\frac{x+1 } {2}\right)}^{\!x+1} \leq x^x$$  is correct.   I’ve tried using mathematical induction methods, but haven’t found a solution. Although, using any method 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.
55741
How one can prove the following.  Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:  $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$  WLOG one can assume that $0\le a\le b\le c$.  It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$.
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?
88080
I would like to prove by induction the following inequality:  $\frac{4^n}{n+1} &lt; \binom{2n}{n}$, for all natural numbers n > 1.  Any hints?
67806
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} &gt; \ln 2$$  $n$ is positive integer.  Thank you !
38991
I find it very strange that $$ e^x \geq x^e \, \quad \forall x \in \mathbb{R}^+.$$   I have scratched my head for a long time, but could not find any logical reason. Can anybody explain what is the reason behind the above inequality? I know this is math community, but I'd appreciate a more intuitive explanation than a technical one.
11088
How do I prove the following inequality  $$\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$$  without the help of induction?  Thanks for any help!!
52673
Does anybody have a simple proof this inequality  $$e^x\le x+e^{x^2}.$$  Thanks.
31850
Show that $$n! \leqslant \left(\frac{n+1}{2}\right)^n \quad \hbox{for all } n \in \mathbb{N}$$  I know that it can be done by induction but I always find line where I do not know what to do next.
89932
For $a,b \in \mathbb R$, $p\geq2$ I try to show $$\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p.$$  Is this a popular inequality (At least I could not find it in the list of popular inequalities from wikipedia)? It seems to be related to convexity but I did not succeed to show it. A related inequality seems to be for $p \geq 1,a,b\geq0$  $$\left(\frac{a+b}{2}\right)^p\leq \frac{1}{2}a^p+\frac{1}{2}b^p,$$ which directly follows from the convexity of $x^p$ for positive numbers.
8096
I am trying to prove that $n^2 \leq 2^n$ for all natural $n$ with $n \ne 3$. My steps are:    induction base case: $n=0:$ $0² \leq 2⁰$ which is okay.    inductive step: $n \rightarrow n+1:$ $(n+1)²\leq2^{n+1}$ $$(n+1)^2 = n^2 + 2n + 1 = ...help...\leq 2^{n+1}$$   I know the bernoulli inequality but don't know where to use it, if I even need to. I have problems when it comes to proving things which are based on orders..
89429
I know from papers like Dusarts's that $$n\left(\ln n +\ln \ln n -1+ \frac{\ln \ln n -2}{\ln n}-\frac{\ln^2 \ln n -6 \ln \ln n +12}{2 \ln^2 n}\right) \leq p_n \leq n\left(\ln n +\ln \ln n -1+ \frac{\ln \ln n -2}{\ln n}-\frac{\ln^2 \ln n -6 \ln \ln n +10}{2 \ln^2 n}\right) $$  But could any one give and explicit number $n_0$ for which for all $n \geq n_0$ the above inequalities hold true ?  It does not have to be the smallest one, also if there is way, could you please show me how to get this number   I read a paper that estimate that the $n_0 \approx 3.9*10^{31}$ under the R.H. if we don't assume R.H. could we still have $n_0 \ll 10^{100}$  Thanks in advance