YanYau wrote: The sequence $\{x_n\}$ is defined by $x_1=5$ and $x_{k+1}=x_k^2-3x_k+3$ for $k=1,2,3\cdots$. Prove that $x_k>3^{2^{k-1}}$ for any positive integer $k$. We can prove by induction that $x_n\geq 3^{2^{n-1}}+2$ as $x_1=5$ Assume that $x_k\geq 3^{2^{k-1}}+2$ for some $k$ $x_{k+1}=x_k(x_k-3)+3\geq (3^{2^{k-1}}+2)(3^{2^{k-1}}-1)+3=3^{2^k}+3^{2^{k-1}}+1>3^{2^{k}}+2$ Done

actually full induction satisfying very well

We claim that $x_k \geq 3^{2^{k-1}}+2$ for all $k \in \mathbb{Z}^+$. We prove by induction. The case $k=1$ is trivial. Assume it is true for some $m \in \mathbb{Z}^+$, then $x_{m+1}=x_m^2-3x_m+3=(x_m-1.5)^2+0.75 \geq (3^{2^{m-1}}+0.5)^2+0.75=3^{2^m}+3^{2^{m-1}}+1 \geq 3^{2^m}+2$, so it is true for $m+1$. Hence we are done.