Prove the inequality $$ 3^{n}\ge2n^{2}+1 $$
for $n=1,2,\dots$
This looks like a problem which might be solved using the binomial theorem. Recall \begin{align*} (1+x)^{n} & =1+nx+{n \choose 2}x^{2}+\dots\\ & =1+nx+\frac{n!}{(n-2)!2!}x^{2}+\dots\\ & =1+nx+\frac{n(n-1)}{2}x^{2}+\dots \end{align*}
Using the binomial theorem to expand the LHS of our original inequality, we have \begin{align*} 3^{n}=(1+2)^{n} & \ge2n^{2}+1\\ 1+2n+\frac{n(n-1)}{2}2^{2}+\dots & \ge1+2n^{2}\\ 1+2n+2n(n-1)+\dots & \ge1+2n^{2}\\ 1+2n+2n^{2}-2n+\dots & \ge1+2n^{2}\\ 1+2n^{2}+\dots & \ge1+2n^{2} \end{align*}
This is obviously true.
Let's try another approach: induction.
Base case
\begin{align*} 3^{0} & \ge2(0)^{2}+1\\ 1 & \ge1 \end{align*}
Now we suppose the normal case $$ 3^{k}\ge2k^{2}+1 $$
For the inductive step, multiple both sides by 3
\begin{align*} 3\cdot3^{k} & \ge3(2k^{2}+1)\\ 3^{k+1} & \ge6k^{2}+3 \end{align*}
So it is sufficient for us to prove that \begin{align*} 6k^{2}+3 & \ge2(k+1)^{2}+1\\ & =2(k^{2}+2k+1)+1\\ & =2k^{2}+4k+3\\ 6k^{2}-2k^{2}+3-3 & \ge4k\\ 4k^{2} & \ge4k \end{align*}
which is obviously true. Thus our original inequality holds
$$ 3^{n}\ge2n^{2}+1. $$
Question: are there problems with either proof or style things that I could learn?
Question: For the inductive proof, did I choose the correct base case of zero? It seemed correct given the context, but I am not 100% certain.
I don't have a solution to check against. Are my proofs valid?
