I am doing an inequality induction question that looks like this:
Prove that $2^n>3n^2$ for $n\geq 8$
So I have done Step $1,2$ but I can't finish step $3$
Step $1$: RTP: $n=8$
LHS=$2^8=256$
RHS=$3(8)^2=192$
Therefore true for $n=8$
Step $2$: Ass $n=k$
$2^k>3k^2$
Step 3: RTP: $n=k+1$
$2^{k+1}>3(k+1)^2$
LHS=$2^k \cdot 2$
$>2 \cdot (3k^2)$
$=6k^2$
New aim-RTP: $6k^2>3 \cdot (k+1)^2$
$6k^2>3k^2+6k+3$
I am stuck here, how do I continue beyond this step?
Notice that $6k^2\ge3(k+1)^2\iff 2k^2\ge(k+1)^2\iff 2k^2\ge k^2+2k+1\iff k^2\ge2k+1$.
Now use your assumption that $k\ge8$, so $k^2\ge 8k$.
Hint: write your statement as
$$2^n - 3n^2 > 0.$$
That way, you can work with the both parts at once (as they are on the same side of the inequality).
So, your final step transforms to
$$6k^2 - 3k^2 -6k - 3 = 3k^2 -6k - 3 = 3 (k^2 - 2k - 1) > 3 (k^2 - 2k - 3) = 3(k-3)(k+1).$$
Since we have assumed that $k \ge 8$...
Why don't you consider the real function $f(x) = 3x^2 -6x - 3$ ?
