Proving inequality: Can we show n^n * (n+1)/2)^2n ≥ (n+1)/2)^3?

[juantheron]

How can we prove $$n^n\cdot \left(\frac{n+1}{2}\right)^{2n}\geq \left(\frac{n+1}{2}\right)^3$$

I did not understand from where i have start.

 

[Evgeny.Makarov]

How can we prove $$n^n\cdot \left(\frac{n+1}{2}\right)^{2n}\geq \left(\frac{n+1}{2}\right)^3$$

At least for $n\ge3$ we have $n>\dfrac{n+1}{2}$, so $n^n>\left(\dfrac{n+1}{2}\right)^3$. Also, $\left(\dfrac{n+1}{2}\right)^{2n}>1$, so it makes the left-hand side even bigger. In general, $n^n$ grows much faster than $n^3$, so it dominates the right-hand side for large $n$.

 

[juantheron]

http://mathhelpboards.com/members/evgeny-makarov/ Thanks, But How i can prove $n^n>n^3$ for $n\geq 3$ without Using Induction.

 

[Greg]

That should be $n^n\ge n^3,n\ge3$. Taking logs base $n$,

$n\ge3$

 

[Evgeny.Makarov]

How i can prove $n^n>n^3$ if $n\geq 3$ without Using Induction.

The general fact is that the function $a^x$ is strictly increasing for $a>1$ (and strictly decreasing if $0<a<1$). Therefore, $n\ge3$ implies $n^n\ge n^3$ (here $a=n$). For integer $n$ is it obvious because $n^n=n^{n-3}n^3\ge n^3$ because $n^{n-3}=\underbrace{n\cdot\ldots\cdot n}_{n-3\text{ times}}$ and $n-3\ge0$. We have a product of numbers, each of which is greater than 1, and then we multiply the result by $n^3$. It boils down to the property $x>1,y>0\implies xy>y$.

 

[juantheron]

Thanks Evgeny and Greg, But i have a doubt , Here $$\displaystyle n^n\cdot \left(\frac{n+1}{2}\right)^n\geq \left(\frac{n+1}{2}\right)^n$$ equality hold for $n=1$

But In your solution you have mention that $n\geq 3$. I did not Understand that.

 

[Evgeny.Makarov]

Here $$\displaystyle n^n\cdot \left(\frac{n+1}{2}\right)^n\geq \left(\frac{n+1}{2}\right)^n$$ equality hold for $n=1$

How does it help prove the original inequality?

But In your solution you have mention that $n\geq 3$. I did not Understand that.

This is a wrong thing not to understand. A reasonable thing is not to get a particular part of a proof that is shown to you. For example, it is OK to ask, "How does this inequality follow from the previous one?" But it does not make much sense not to understand something that has not been proved.

I indeed omitted the case when $n< 3$. It is trivial to manually check that the inequality holds for those values of $n$. With some thought it may be possible to construct a single general proof, which does not make a distinction whether $n\ge 3$ or not. Indeed, $\left(\frac{n+1}{2}\right)^{2n}\geq \left(\frac{n+1}{2}\right)^3$ for all $n\ge1$ because for $n=1$ both sides equal 1, but for $n\ge2$ we have $2n>3$.

 

[juantheron]

Thanks evgeny Now i have got it.

Yes you are saying Right actually i have a problem for $n < 3$

and I appolozise for using such word.