Stuck in Induction Inequality: $2^n>3n^2$ I am doing an inequality induction question that looks like this:

Prove that $2^n&gt;3n^2$ 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 $n=k$

$2^k&gt;3k^2$

Step 3: RTP: $n=k+1$

$2^{k+1}&gt;3(k+1)^2$

LHS=$2^k \cdot 2$

$&gt;2 \cdot (3k^2)$

$=6k^2$

New aim-RTP: $6k^2&gt;3 \cdot (k+1)^2$

$6k^2&gt;3k^2+6k+3$

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 $k^2\ge 8k$.
Why don't you consider the real function $f(x) = 3x^2 -6x - 3$ ?