$x_1=\sqrt{2}, x_{n+1}=x_n+\frac 1 {x_n}$ then show that $\displaystyle \sum_{i=1}^{2021} \dfrac {x_i^2}{2x_ix_{i+1}-1}>\dfrac{2021^2}{x_{2021}^2+\dfrac 1 {x_{2021}^2}}.$
My Attempt:
\begin{align} &2x_ix_{i+1}-1=2x_i\bigg(x_i+\frac 1 {x_i} \bigg)-1=2x_i^2+1. \\ &\therefore \sum_{i=1}^{2021} \frac {x_i^2}{2x_1x_{i+1}-1} = \sum_{i=1}^{2021} \frac{x_i^2}{2x_i^2+1} = \frac 1 2 \sum_{i=1}^{2021} 1-\frac 1 {2x_i^2+1} = \frac {2021}{2}-\sum_{i=1}^{2021}\frac 1 {4x_i^2+2}. \\ &x_{2021}^2+\dfrac {1} {x_{2021}^2} = \bigg(x_{2021}+\dfrac 1 {x_{2021}} \bigg)^2-2 = x_{2022}^2-2. \\ \ \\ &\therefore ETS) \ \dfrac {2021} {2} - \sum_{i=1}^{2021} \dfrac 1 {4x_i^2+2} > \dfrac {2021^2}{x_{2022}^2-2}. \end{align}
The OP wrote that $\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} = \sum_{i=1}^{2021} \dfrac{x_i^2}{2x_i^2+1} \ $
If $f(x)=\dfrac{1}{x}$, we can write that
$\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} = \sum_{i=1}^{2021} f\left( 2+\dfrac{1}{x_i^2}\right) = \sum_{i=1}^{2021} f\left( x_{i+1}^2-x_i^2\right)$
$f$ is strictly convex on $(0,+\infty )$ because $f''$ is positive, then:
$\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} > 2021 f\left( \dfrac{\sum_{i=1}^{2021} (x_{i+1}^2-x_i^2)}{2021}\right) = 2021 f\left( \dfrac{x_{2022}^2-2}{2021}\right)$
So we get
$\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} > \dfrac{2021^2}{x_{2022}^2- 2} = \dfrac{2021^2}{x_{2021}^2+\dfrac{1}{x_{2021}^2}}$