$LHS = n + \sum \limits_{i < j} (\frac{a_i}{a_j}+\frac{a_j}{a_i})$ Given $a_p \neq a_q$, we have that $(\frac{a_r}{a_p}+\frac{a_p}{a_r})+(\frac{a_r}{a_q}+\frac{a_q}{a_r}) = a_r (\frac{a_p+a_q}{a_pa_q}) + \frac{a_p+a_q}{a_r} \geq \frac{2(a_p+a_q)}{\sqrt{a_pa_q}}$ which is some finite constant depending on $a_p, a_q$ strictly greater than 4, call it $\epsilon$. It follows that $LHS = n + \sum \limits_{i < j, i, j \neq p, q} (\frac{a_i}{a_j}+\frac{a_j}{a_i}) + \sum \limits_{r \neq p, q} (\frac{a_r}{a_p}+\frac{a_p}{a_r}+ \frac{a_r}{a_q}+\frac{a_q}{a_r}) + \frac{a_p}{a_q}+\frac{a_q}{a_p} > n + (n-2)(n-3) + (n-2)\epsilon + 2 = n^2 + (\epsilon - 4)n+ (8-2\epsilon) \leq n^2+2019$ $(\epsilon - 4)n < (2\epsilon - 8)+2019$. Since $a_p \neq a_q$ for all n above the current value of n, we may increase $n$ as we please. This leads to a contradiction, as simply setting $n = \left\lceil \frac{2\epsilon + 2011}{\epsilon - 4} \right\rceil$ suffices. This means for no value of $n$ can $a_p \neq a_q$, thus $\{a_n\}$ is constant. EDIT: Did not read that the sequence is natural, this solves for all positive reals. Hopefully, it still stands true. EDIT 2: It probably still does work since the inequality is homogenous, the natural case generalizes easily to the rational case, which is basically synonymous with real ineqs.

Nice Solution! Here's another one using AM-HM inequality. $pf)$ Dividing both sides in $n^2$, we obtain $$\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)\left(\frac{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}{n}\right)\le{1+\frac{2019}{n^2}}$$and since the inequality holds for all natural number $n$, we can say that $LHS=1$ for big enough $n$. By AM-HM inequality, $LHS\ge1$, and equality holds when $a_{1}=\cdots=a_{n}$ which means $\{a_{n}\}$ is constant!

I shall add above to my collection "classic fakesolves".

Rama12 wrote: Nice Solution! and since the inequality holds for all natural number $n$, we can say that $LHS=1$ for big enough $n$. By AM-HM inequality, $LHS\ge1$, and equality holds when $a_{1}=\cdots=a_{n}$ which means $\{a_{n}\}$ is constant! How can you just say that $LHS = 1$?, it only is approximately $1$ and u can't use the fact that $LHS \ge 1$ to solve it

Well, $$\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)\left(\frac{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}{n}\right)\le{1+\frac{2019}{n^2}}$$$$\lim_{n\to\infty}\biggl\{\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)\left(\frac{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}{n}\right)\biggl\}\le\lim_{n\to\infty}\left({1+\frac{2019}{n^2}}\right)=1$$so, $LHS=1$ for big enough $n$. Am I wrong? I don't know much about the limit...

Adapted (false) problem wrote: $\{a_{n}\}$ is a sequence of natural numbers satisfying the following inequality for all natural number $n$: $$(a_{1}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)\le{n^{2}}+2019n$$Prove that $\{a_{n}\}$ is constant. Adapted (fake) solution wrote: Dividing both sides in $n^2$, we obtain $$\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)\left(\frac{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}{n}\right)\le{1+\frac{2019}{n}}$$and since the inequality holds for all natural number $n$, we can say that $LHS=1$ for big enough $n$. By AM-HM inequality, $LHS\ge1$, and equality holds when $a_{1}=\cdots=a_{n}$ which means $\{a_{n}\}$ is constant!

Rama12 wrote: Well, $$\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)\left(\frac{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}{n}\right)\le{1+\frac{2019}{n^2}}$$$$\lim_{n\to\infty}\biggl\{\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)\left(\frac{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}{n}\right)\biggl\}\le\lim_{n\to\infty}\left({1+\frac{2019}{n^2}}\right)=1$$so, $LHS=1$ for big enough $n$. Am I wrong? I don't know much about the limit... Just because the limit equals $1$ does not mean that it is equal to exactly $1$ for large $n$. Also, the fact that $LHS > 1$ for large $n$ if the sequence is not constant still does not imply $LHS = 1$ for large $n$. Take for example $\frac{1}{n}$ and $\frac{2}{n}$. Both converges to $0$ as $n$ approaches infinity, but $\frac{1}{n} < \frac{2}{n}$ for all $n > 0$. Anyway, here's my solution (I repeatedly got LaTeX errors when typing this it's so annoying lol): Assume the contrary, then there exists $i > j$ such that $a_i \neq a_j$. Let $x_n = (a_{1}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)-{n^{2}}$. It is given that $x_n \leq 2019$ and by Cauchy-Schwarz, $x_n \geq 0$ for all $n$. We will prove that there exists a positive constant $c$ such that $x_{n+1} - x_n \geq c$ for all $n \geq i$. Indeed, for $n \geq i$, we have \begin{align*} x_{n+1} - x_{n} &= (a_{1}+\cdots+a_{n+1})\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n+1}}\right) - (a_{1}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right) - (2n + 1) \\ &= (a_{1}+\cdots+a_{n})\frac{1}{a_{n+1}} + \left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right) a_{n+1} - 2n \\ &= \sum_{k = 1}^{n} (\frac{a_k}{a_{n+1}} + \frac{a_{n+1}}{a_k}) - 2n \\ &\geq \left(\frac{a_i}{a_{n+1}} + \frac{a_{n+1}}{a_i}\right) + \left(\frac{a_j}{a_{n+1}} + \frac{a_{n+1}}{a_j}\right) - 4 \\ &= (a_i + a_j)\left(\frac{1}{a_{n+1}} + \frac{a_{n+1}}{a_ia_j}\right) - 4 \\ &\geq \frac{2(a_i + a_j)}{\sqrt{a_ia_j}} - 4 \\ &= \frac{2(\sqrt{a_i} - \sqrt{a_j})^2}{\sqrt{a_ia_j}} \end{align*}which is a positive constant. Hence, $x_{i+k} \geq x_{i+k} - x_i \geq kc$ for all positive integer $k$, and taking $k$ sufficiently large we find $x_n > 2019$, a contradiction.

@hyay hyay wrote: Just because the limit equals $1$ does not mean that it is equal to exactly $1$ for large $n$. Also, the fact that $LHS > 1$ for large $n$ if the sequence is not constant still does not imply $LHS = 1$ for large $n$. Take for example $\frac{1}{n}$ and $\frac{2}{n}$. Both converges to $0$ as $n$ approaches infinity, but $\frac{1}{n} < \frac{2}{n}$ for all $n > 0$. Thanks for your advice!

NTstrucker wrote: Adapted (false) problem wrote: $\{a_{n}\}$ is a sequence of natural numbers satisfying the following inequality for all natural number $n$: $$(a_{1}+\cdots+a_{n})\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)\le{n^{2}}+2019n$$Prove that $\{a_{n}\}$ is constant. Adapted (fake) solution wrote: Dividing both sides in $n^2$, we obtain $$\left(\frac{a_{1}+\cdots+a_{n}}{n}\right)\left(\frac{\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}{n}\right)\le{1+\frac{2019}{n}}$$and since the inequality holds for all natural number $n$, we can say that $LHS=1$ for big enough $n$. By AM-HM inequality, $LHS\ge1$, and equality holds when $a_{1}=\cdots=a_{n}$ which means $\{a_{n}\}$ is constant! Oh, I understood. Thanks for explaining it.

Is it true ? Let $A = (\sum a_{i})(\sum \frac{1}{a_{i}})-n^2$ . It is easy to see $$A=(\sum \sqrt{a_{i}}^2)(\sum \sqrt{\frac{1}{a_{i}}}^2)-(\sum \sqrt{a_{i}}^2.\sqrt{\frac{1}{a_{i}}}^2) = \sum_{i \neq j} (\frac{\sqrt{a_i}}{\sqrt{a_j}}-\frac{\sqrt{a_j}}{\sqrt{a_i}})^2 =\sum_{i \neq j} \frac{(a_{i}-a_{j})^2}{a_{i}a_{j}}$$Also we can see $\sum_{i \neq j} \frac{(a_{i}-a_{j})^2}{a_{i}a_{j}} \geq \sum_{i \neq j} \frac{|a_{i}-a_{j}|}{a_{i}a_{j}} = \sum_{i \neq j} |\frac{1}{a_i}-\frac{1}{a_j}|$ ( ) We can get $ |\frac{1}{a_i}-\frac{1}{a_j}| \geq \frac{1}{k}-\frac{1}{k+1} = \frac{1}{k(k+1)}$ . ( ) By ( ) and ( ) we can see $$A \geq n^2+\frac{n}{k(k+1)}$$( ) In ( ) is not true for all $n$ $$n \geq 2020 k(k+1)$$Because if different $n$ apply in the above inequality, this inequality is false . We can see $\{a_{n}\}$ is constant. $\blacksquare$

@above that is a nice other way to look at this problem (but whats up with the emojis lol)

Let's think that there are $i$ and $j$ which follows $ {a}_{i} \neq {a}_{j} $ $ n^2+2019 \ge ({a}_{1}+\cdot \cdot \cdot +{a}_{i}+\cdot \cdot \cdot +{a}_{j}+\cdot \cdot \cdot +{a}_{n})(\frac{1}{{a}_{1}}+ \cdot \cdot \cdot +\frac{1}{{a}_{j}}+\cdot \cdot \cdot +\frac{1}{{a}{i}}+\cdot \cdot \cdot +\frac{1}{{a}_{n}}) $ $ \ge $ $ (n-2+\sqrt{\frac{{a}_{j}}{{a}_{i}}}+\sqrt{\frac{{a}_{i}}{{a}_{j}}} )^2 $ $ (c \ge 0) $ but this means that this works only when $ n \le \frac{2019-c^2}{2c} $ , but it should work for all n. So, all $ {a}_{i} $ are the same.