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?

  • 1
    $\begingroup$ your proofs look basically correct; you could choose base case $0$ to prove the proposition for $0, 1, 2, ...$ or base case $1$ to prove it for $1, 2, ...$ $\endgroup$ – J. W. Tanner Sep 24 '19 at 2:18
  • $\begingroup$ How and why do you choose 0 or 1? For me, I chose 0 because there didn't seem to be any stated constraints on n, say by another inequality, and we are considering the natural numbers in induction. Is that proper reasoning? $\endgroup$ – Joe Sep 24 '19 at 2:20
  • 1
    $\begingroup$ you could choose whichever you like if it works, but if you're asked to prove something for $1,2,...$, there's no need to prove it for $0$ $\endgroup$ – J. W. Tanner Sep 24 '19 at 2:21
  • $\begingroup$ I forgot to consider the problem statement of domain for n. Sorry for the dumb question. $\endgroup$ – Joe Sep 24 '19 at 2:22

Your proofs look basically correct. When you did the proof by induction, you took the base case as $0$; you could have merely started with the base case of $1$, since the question asked to prove the inequality only for $n=1,2,...$.


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