a)By AM-GM ineaquality $a_{n+1}=\frac{a_n^3}{16}+\frac{3}{16a_n}+\frac{3}{16a_n}+\frac{3}{16a_n}$ $\geq \sqrt [4] {\frac{a_n^3}{16} \frac{3}{16a_n} \frac{3}{16a_n} \frac{3}{16a_n}}>$ $>\frac{4}{5}$ By I dot know $a_{n+1}<\frac{5}{4}$

So nobody notices that $a_1 = 2 > \frac {5} {4}$ and that $a_2 = \frac {25} {32} < \frac {4} {5}$, so none of those inequalities is valid for all $n$ ? So by your AM-GM inequality $a_{n+1}=\frac{a_n^3}{16}+\frac{3}{16a_n}+\frac{3}{16a_n}+\frac{3}{16a_n}$ $\geq 4\sqrt [4] {\frac{a_n^3}{16} \frac{3}{16a_n} \frac{3}{16a_n} \frac{3}{16a_n}}<$ $\frac{4}{5}$ ? (I made two corrections, one putting the forgotten factor $4$, second, reversing the last inequality to its true value) ? Good work ... No mention about the only correct thing that is easy to prove, namely that if $a_n < \frac {5} {4}$, then $a_{n+1} < \frac {5} {4}$ ...

I think, we can not find the value of $a_n \ge 2$, because the function hold for $n \ge 2$ not $n \ge 1$. And, from mavropnevma post. So, i think this problem totally wrong.

Here is the actual problem Source- Inequalities- An Approach Through Problems by BJ Venkatachala Quote: Define a sequence $\{x_n\}$ by $$x_1=2,x_{n+1}=\frac{x_n^4+9}{10x_n}.$$Prove that $\frac45<x_n\leq\frac54$ for all $n>1$

Here's the solution By simple AM-GM, $$\frac{x_n^3}{10}+\frac{3}{10x_n}+\frac{3}{10x_n}+\frac{3}{10x_n}\geq\frac25\sqrt[4]{27}>\frac45$$So, $x_n>\frac45$ for all $n>1$ We have $x_2=\frac54$ It suffices to prove that the sequence is decreasing, i.e., $$\frac{x_n^4+9}{10x_n}\leq x_n$$Equivalently we need to prove $$(x_n^2-1)(x_n^2-9)\leq 0$$If $x_n\geq 1$, the inequality is obviously true. If $x_n<1$, $$\frac{x_n^4+9}{10x_n}<\frac{1}{x_n}<\frac54$$from first part. So, $\frac45<x_n\leq \frac54$ as required