First, we will prove the following lemma. Lemma 01. For any positive integer $k$, we have \[ x_{k + 1}^2 = 2k + 2 + \sum_{n = 1}^k \frac{1}{x_n^2} \]Proof. We will prove this by induction. The case $k = 1$ is true, as \[ x_2^2 = x_1^2 + \frac{1}{x_1^2} + 2 = 4 + \frac{1}{x_1^2} \]Now, suppose that the statement is true for $k = m$, then \[ x_{m + 2}^2 = \left( \frac{1}{x_{m+1}^2} + 2 \right) + x_{m+1}^2 = \left( \frac{1}{x_{m+1}^2} + 2 \right) + 2(m + 1) + \sum_{n = 1}^m \frac{1}{x_n^2} = 2(m+2) + \sum_{n = 1}^{m+1} \frac{1}{x_{n}^2} \]Hence, the statement itself is true by induction. Now, to finish the problem, rewrite the inequality as: \begin{align*} \sum_{k = 1}^{2019} \frac{x_k^2}{2x_kx_{k + 1} - 1} &= \sum_{k = 1}^{2019} \frac{x_k^2}{2x_k^2 + 1} \\ &= \sum_{k = 1}^{2019} \frac{1}{2 + \frac{1}{x_k^2}} \\ &> \frac{2019^2}{4038 + \sum_{k = 1}^{2019} \frac{1}{x_k^2}} \\ &= \frac{2019^2}{x_{2020}^2 - 2} \\ &= \frac{2019^2}{x_{2019}^2 + \frac{1}{x_{2019}^2} } \end{align*}by CS Engel and $\textbf{Lemma 01.}$

ThE-dArK-lOrD wrote: Let $(x_n)_{n\in \mathbb{N}}$ be a sequence defined recursively such that $x_1=\sqrt{2}$ and $$x_{n+1}=x_n+\frac{1}{x_n}\text{ for }n\in \mathbb{N}.$$Prove that the following inequality holds: $$\frac{x_1^2}{2x_1x_2-1}+\frac{x_2^2}{2x_2x_3-1}+\dotsc +\frac{x_{2018}^2}{2x_{2018}x_{2019}-1}+\frac{x_{2019}^2}{2x_{2019}x_{2020}-1}>\frac{2019^2}{x_{2019}^2+\frac{1}{x_{2019}^2}}.$$ Proposed by Ivan Novak

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IndoMathXdZ wrote: First, we will prove the following lemma. Lemma 01. For any positive integer $k$, we have \[ x_{k + 1}^2 = 2k + \sum_{n = 1}^k \frac{1}{x_n^2} \] It's wrong. Try $k=1$.

\[ x_{k }^2 = 2k + \sum_{n = 1}^{k-1} \frac{1}{x_n^2} \]

Very nice problem for a P2, now I present my solution which doesn't use induction and it's very simple. Solution. Squaring $x_{n+1}=x_n+\frac{1}{x_n}\implies x_{n}^2+\frac{1}{x_{n}^2}=x_{n+1}^2-2...(\clubsuit)$, now $RHS=\frac{2019^2}{x_{2020}^2-2}$. Also, notice that $2x_nx_{n+1}-1=2x_n(x_n+\frac{1}{x_n})-1=2x_n^2+1$, now we have that $$\frac{x_n^2}{2x_1x_2-1}=\frac{x_n^2}{2x_n^2+1}=\frac{x_n^2}{x_n^2(2+\frac{1}{x_n^2})}=\frac{1}{2+\frac{1}{x_n^2}}\stackrel{\clubsuit}{=}\frac{1}{x_{n+1}^2-x_{n}^2}$$$$LHS=\sum_{n = 1}^{2020}\frac{1}{x_{n+1}^2-x_{n}^2}\stackrel{AM-HM}{\geq}\frac{2019^2}{(x_{2}^2-x_{1}^2)+...+(x_{2020}^2-x_{2019}^2)}=\frac{2019^2}{x_{2020}^2-x_{1}^2}=\frac{2019^2}{x_{2020}^2-2}=RHS$$It is clear that the equality cannot hold.$\blacksquare$

Beautiful problem