An Inequality with increasing $3$'s and a Fibonacci sequence.

Question

Prove that :- $\frac{\sqrt{n+1}}{n} + \frac{\sqrt{n+4}}{n+3} + \frac{\sqrt{n+7}}{n+6} + \frac{\sqrt{n+10}}{n+9} + \frac{\sqrt{n+13}}{n+12} < \frac{1}{\sqrt{n-1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+5}} + \frac{1}{\sqrt{n+8}} + \frac{1}{\sqrt{n+11}}$ .

What I Tried: I first think that the Right-Hand-Side should contain, $\Big(\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+5}} + \frac{1}{\sqrt{n+8}} + \frac{1}{\sqrt{n+11}}\Big)$ instead of that $(n-1)$ , so I think that is a typo given in the question and that can be ignored.

Other than that, as the title suggests, I can see that the left-hand-side has the numerator and the denominator to be increasing by $3$ , and the terms of the denominator in the right-hand-side, form the fibonacci sequence.
So an usual question comes, can this inequality be generalised to more terms?

As a try for solving it, as I am a novice in Inequality I tried AM-GM and C-S , but they definitely won't work here I guess, because the Inequality seems a bit unusual. Another idea was that I thought of proving each separate term is greater than it's corresponding term in the right hand side, forming a bijection. But that did not work either, as I saw $\frac{\sqrt{n+1}}{n} > \frac{1}{\sqrt{n-1}}$ .

Can anyone help me here? Thank You.

Answer 1

The inequality is OK, there is no typo.

BlockquoteI thought of proving each separate term is greater than it's corresponding term in the right hand side, forming a bijection. But that did not work either, as $\frac{\sqrt{n+1}}{n} > \frac{1}{\sqrt{n-1}}$.

This luckily is not true. We have $$\frac{\sqrt{n+1}}{n} < \frac{1}{\sqrt{n-1}},$$ as $$\sqrt{n+1}\sqrt{n-1} < n,$$ $$\sqrt{n^2-1}<n.$$

Summing up, we get the inequality.

Answer 2

That is not a typo. If you think that the inequality holds with $n+1$, it should hold with $n-1$ too (why?). Also, there is nothing related to Fibonacci series here. The Fibonacci series is $1,1,2,3,5,8,13,21,\dots$.

Recheck your claim that $\frac{\sqrt{n+1}}{n}>\frac1{\sqrt{n-1}}$. Its incorrect. Try substituting $n=2$ or $n=3$ and see it for yourself.

Another idea was that I thought of proving each separate term is greater than it's corresponding term in the right hand side.

This idea does work indeed. Continue your approach. Try proving that \begin{align*} \frac{\sqrt{n+i+2}}{n+i+1}<\frac1{\sqrt{n+i}} \end{align*} for any $i$ for which the denominators are positive and square roots are well defined.