$a_1=3$ and $a_{n+1}=\dfrac{2a_n}{3}+\dfrac{4}{3a_n^2}$. Show that $4^{1/3} \le a_n$for all $1\le n$.

Question

$a_1=3$ and $a_{n+1}=\dfrac{2a_n}{3}+\dfrac{4}{3a_n^2}$

By considering the function $f(x)=\dfrac{2x}{3}+\dfrac{4}{3x^2}$, show that $4^{1/3} \le a_n$ for all $1\le n$.

Answer 1

$f'(x)=\dfrac{2}{3}-\dfrac{8}{3x^3}=0 \implies x=4^{\frac{1}{3}}$ it is easy to verify this is min point when$x>0$ ,so $f_{min}=4^{\frac{1}{3}}$

$a_n$ is belong to $f(x)$ so $a_n \ge 4^{\frac{1}{3}}$