(Variation) Let $a_1 >0$ and \begin{align*} a_{n+1}=\frac{a_n}{1+a_n^2} \end{align*}for $n \geq 1$. Prove that there exists an $n$ such that $10 \sqrt{n} a_n >7$

Bumping.

Rewriting the characteristic equation as $\frac{a_n}{a_n+1}=1+a_n^2$ Note, $$\prod_{i=1}^n\frac{a_i}{a_{i+1}}=\prod_1^n(1+a_i^2)\to\frac {1}{a_{n+1}}=\prod_1^n\,(1+a_i^2)\,(\clubsuit)$$Suppose that $\forall\,a_i\,i=\overline{1,n},$ with enough large $n.$ Assume that $a_i\le\frac{7}{10\sqrt n}.$ $$(\clubsuit)\implies\frac{1}{a_i+1}\le\prod_1^n\left(1+\frac{49}{100n}\right)=e^{\sum_1^n\ln\left(1+\frac {49}{100n}\right)}\,(\spadesuit)$$Applying the inequality $\ln(1+x)<x\,, x>0$ in each factor of $(\spadesuit),$ we obtained $\frac{1}{a_{n+1}}< e^{\frac{49}{100}\sum_1^n\tfrac 1n}$ If there exists a number $n+1\implies\,10 (n+1)^{\tfrac 12}a_{n+1}>7\to\frac{1}{a_{n+1}}<\frac{10\sqrt{n+1}}{7}$ as a consequence. We will prove the inequality $$e^{\frac{49}{100}\sum_n^1\tfrac 1n}<\frac {10\sqrt{n+1}}{7}\Leftrightarrow\frac{49}{100}\sum_1^n\tfrac 1n <\ln\left(\frac{10}{7}\right)+\tfrac 12\ln(n+1)\,(\diamondsuit) $$We will prove stronger result $$\sum_1^n\tfrac 1n< 1+\ln(n+1)+\left(\frac{1}{50}\sum\tfrac 1n-(1-2\ln(\frac{10}{7}))\right)\,(\star) $$$(\star) $ holds for the integral inequality. Note that, the sum of harmonic series $\{\tfrac 1n\}\mapsto\infty.$ We can take sufficiently large $n $ to obtain the positive value of latter part in $\text{RHS}.$