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juantheron
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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.
I did not understand from where i have start.
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 $n$.jacks said:How can we prove $$n^n\cdot \left(\frac{n+1}{2}\right)^{2n}\geq \left(\frac{n+1}{2}\right)^3$$
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 $x>1,y>0\implies xy>y$.jacks said:How i can prove $n^n>n^3$ if $n\geq 3$ without Using Induction.
How does it help prove the original inequality?jacks said:Here \(\displaystyle \displaystyle n^n\cdot \left(\frac{n+1}{2}\right)^n\geq \left(\frac{n+1}{2}\right)^n\) equality hold for $n=1$
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.jacks said:But In your solution you have mention that $n\geq 3$. I did not Understand that.
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