# Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$

Prove that for all $$n \in \mathbb{N}$$ the inequality $$\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$$ holds.

My work. I proved this inequality, but my proof is ugly (it is necessary to check by brute force whether the inequality holds for $$n=1,2,3,4,5$$).

• Have you looked at Riemann sums ( en.wikipedia.org/wiki/Riemann_sum ) ? – user7440 Aug 2 '19 at 21:03
• $\sum_{n=1}^{N} f(n) = \int_{\varepsilon}^{N+1} f(x)d\lfloor x \rfloor, ~ 0<\varepsilon<1$ – Luis Felipe Aug 2 '19 at 21:08
• The first expression looks like $\approx n^2-(log(n))^2\lt \frac{9n^2}{8}$. – herb steinberg Aug 2 '19 at 21:40

First problem.

Let $$x_k=\frac{k+1}{k}.$$

Thus, $$x_k\in[1,2]$$ for all $$k$$.

Now, let $$f(x_1,x_2,...x_n)=\sum_{k=1}^nx_k\sum_{k=1}^n\frac{1}{x_k}.$$ Easy to see that $$f$$ is a convex function of $$x_k$$ for all $$k$$.

Thus, by AM-GM $$\max_{x_k\in[1,2]}f=\max_{x_k\in\{1,2\}}f=\max_{i+j=n}(i+2j)\left(i+\frac{j}{2}\right)=$$ $$=\max_{i+j=n}\left((i+j)^2+\frac{1}{2}ij\right)\leq n^2+\frac{1}{2}\left(\frac{n}{2}\right)^2=\frac{9}{8}n^2.$$

• I'm very interested in your proof, but unfortunately I do not understand your proof. Please explain this step $$\max_{x_k\in[1,2]}f=\max_{x_k\in\{1,2\}}f=\max_{i+j=n}(i+2j)\left(i+\frac{j}{2}\right)$$ in more detail. Also i don't know what is AM-GM. – Witold Aug 3 '19 at 9:00
• @Witold Convex function gets a maximal value for extreme value of the variable. Draw it! For example if you'll take a parabola $y=x^2$ on $[-1,2]$ so since $y=x^2$ is a convex function $x^2$ gets a maximal value for $x=-1$ or for $x=2$, which gives $\max_{[-1,2]}x^2=\max\{(-1)^2,2^2\}=4.$ In our case it's enough to prove our inequality for $x_k\in\{1,2\}$. AM-GM here it's the following: $ij\leq\left(\frac{i+j}{2}\right)^2.$ – Michael Rozenberg Aug 3 '19 at 9:09
• Thank you! I understand what you wrote. But how to prove that the function $f$ is convex? – Witold Aug 3 '19 at 9:40
• @Witold I did not say that $f$ is a convex function. I said that $f$ is a convex function of $x_k$ for all $k$. I see it so. Let $x_k=x$. Thus, $f=(x+A)\left(\frac{1}{x}+B\right)=1+AB+Bx+\frac{A}{x},$ which is a convex function of $x$. – Michael Rozenberg Aug 3 '19 at 9:46
• Very nice proof! Thank you very much! – Witold Aug 3 '19 at 9:55

The first inequality is equivalent to $$(n+H_n)(n+1-H_{n+1})\le\frac98n^2$$ where $$H_n$$ denotes the $$n$$th Harmonic number. It's well known that $$H_n=\gamma+\ln{(n)}+o(1)$$ as $$n\to\infty$$ so it's clear that the LHS is asymptotic to $$n^2+n+o(n)$$. In particular, we have that the LHS divided by $$n^2$$ is decreasing. So we just need to find the ratio of this function and $$n^2$$ for small $$n$$ and as the ratio begins to decrease, it will not increase again due to the function's asymptotic behaviour. Using this gives us the much stricter inequality $$(n+H_n)(n+1-H_{n+1})\le\frac{104273}{100800}n^2$$ for $$n\in\mathbb{N}$$. Equality only occurs when $$n=6$$ and the ratio is at a maximum.

The second inequality is equivalent to $$(n^2+2n-H_n)(n^2-2n-3+3H_{n+1})\le\frac98n^4$$ A similar analysis to before gives us the asymptotic $$n^4+n^2(2\ln{(n)}+2\gamma-7)+o(n^2)$$ for the LHS. Dividing this by $$n^4$$ gives a decreasing ratio so we again only need to analyse small values of $$n$$. This gives us the stricter inequality $$(n^2+2n-H_n)(n^2-2n-3+3H_{n+1})\le\frac{1975}{1944}n^4$$ for $$n\in\mathbb{N}$$. Equality only occurs when $$n=3$$ in this case.

• You can use $H_n = \ln(n)+\gamma+\frac{c}{2k}$ where $0 < c < 1$. – marty cohen Aug 2 '19 at 22:35
• @martycohen I see that it's more useful to use $o(1)$ in this case to prevent the problem from becoming overcomplicated. – Peter Foreman Aug 3 '19 at 9:49
• Thank you very much! – Witold Aug 3 '19 at 10:04