Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Help me to prove the following inequality(2)) without using Bernoulli's inequality. Prove that:

1) .$$\prod_{k=1}^{n}{\sqrt[k+1]{k}} \leq \frac{2^n}{n+1}.$$ 2) $$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2},$$

$n \neq 0 , n \in \mathbb{N}. $

My proof:

1) $$\prod_{k=1}^{n}{\sqrt[k+1]{k}} = \prod_{k=1}^{n}{\sqrt[k+1]{k\cdot 1 \cdot 1 \ldots \cdot 1}} \leq \prod^{n}_{k=1}{\frac{k+1+1+\ldots +1}{k+1}}=$$ $$=\prod^{n}_{k=1}{\frac{2k}{k+1}}=2^n \cdot \prod^{n}_{k=1}{\frac{k}{k+1}}=\frac{2^n}{n+1}.$$

2) $$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq n \cdot \sqrt[n]{\prod_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}}} \geq n \cdot \sqrt[n]{\frac{n+1}{2^n}}=\frac{n}{2}\cdot \sqrt[n]{n+1}.$$

It's remains to prove that

$$\frac{n}{2}\cdot \sqrt[n]{n+1} \geq \frac{n^2+3n}{2n+2}=\frac{n(n+3)}{2(n+1)},$$

or

$$\sqrt[n]{n+1} \geq \frac{n+3}{n+1}$$ or

$$(n+1) \cdot (n+1)^{\frac{1}{n}} \geq n+3.$$

We apply Bernoulli's Inequality and we have:

$$(n+1)\cdot (1+n)^{\frac{1}{n}}\geq (n+1) \cdot \left(1+n\cdot \frac{1}{n}\right)=(n+1)\cdot 2 \geq n+3,$$ which is equivalent with: $$2n+2 \geq n+3,$$ or

$$n\geq 1,$$ and this is true becaue $n \neq 0$, $n$ is a natural number.

Can you give another solution without using Bernoulli's inequality.

Thanks :-)

share|improve this question
1  
Are you sure that $(1+n)^{1/n}\geqslant2$? –  Did Jan 17 '13 at 7:44
    
Good point (+1) –  Chris's sis Jan 17 '13 at 8:01
    
@did yes you are right. It's a bad mistake. Can you give another solution and can you tell me please why Bernoulli's inequality is satisfied? thanks :) –  Iuli Jan 17 '13 at 16:45
add comment

2 Answers

up vote 1 down vote accepted

The inequality to be shown is $$(n+1)^{n+1}\geqslant(n+3)^n, $$ for every positive integer $n$.

For $n = 1$ it is easy. For $n \ge 2$, apply AM-GM inequality to $(n-2)$-many $(n+3)$, 2 $\frac{n+3}{2}$, and $4$, we get $$(n+3)^n < \left(\frac{(n-2)(n+3) + \frac{n+3}{2} + \frac{n+3}{2} + 4}{n+1}\right)^{n+1} = \left(n+1\right)^{n+1}$$

share|improve this answer
add comment

The inequality to be shown is $$(n+1)^{n+1}\geqslant(n+3)^n, $$ for every positive integer $n$. Introduce the function $u$ defined by $$ u(x)=(x+1)\log(x+1)-x\log(x+3), $$ then standard computations yield $$ u'(x)=\frac3{3+x}+\log\left(\frac{1+x}{3+x}\right),\qquad u''(x)=\frac{3-x}{(1+x)(3+x)^2}, $$ hence $u'$ increases on $(0,3)$ and decreases on $(3,+\infty)$. Since $u'(1)=\frac34-\log2\gt0$ and $u'(+\infty)=0$, $u$ is increasing on $(1,+\infty)$. Since $u(1)=0$, this yields $u(n)\geqslant0$ for every positive integer, QED.

Edit: Equivalently, one wants to prove that $(k+2)^{k-1}\leqslant k^k$ for every $k\geqslant2$, that is, $k+2\geqslant\left(1+\frac2k\right)^k$. If one knows that the RHS is increasing and converges to $\mathrm e^2\lt8$, this yields the result for every $k\geqslant6$. The cases $2\leqslant k\leqslant5$ can be checked manually.

share|improve this answer
    
Re Bernoulli's inequality $(1+x)^r\gt1+rx$, recall that it is valid for every nonnegative integer $r$ and every real number $x\geqslant-1$. Using it when $r$ is not an integer can lead to chaos. –  Did Jan 17 '13 at 18:56
    
Yes, it's a nice solution, but I ask you something please. Another solution, it must be a solution for a student who didn't learn derivatives. Thanks :) –  Iuli Jan 17 '13 at 19:50
1  
Don't you think it could have been a good idea to make precise this request BEFORE, i.e. when you asked the question? –  Did Jan 17 '13 at 19:58
    
I asked if you can. I will appreciate your effort, if you don't know or you don't want it's not any problem. In our school, when we see an inequality we don't think to derivatives because these inequalities aren't so hard such that to use derivatives and this inequality must be solved by a children of 15 years old who don't know analysis. Thanks for your help, I appreciate your answers - are very clearly and useful, again thanks :) –  Iuli Jan 17 '13 at 20:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.