# Stuck in Induction Inequality

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?

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why are you writing the product as . instead of $\cdot$? –  Alex Jan 18 '14 at 23:32
"\cdot" is the symbol that accomplishes this, BTW –  MPW Jan 18 '14 at 23:44
You should take a look here to learn how to properly format math in LaTeX. –  Olivier Jan 18 '14 at 23:48

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 $k^2\ge 8k$.

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is there a way to prove $k^2>2k+1$ without using the assumption? –  user117196 Jan 18 '14 at 23:48
@user117196 For $k = 0$ or $k = 1$, this is not true. Hence, you need some assumption on $k$. –  Vedran Šego Jan 18 '14 at 23:50

$$2^n - 3n^2 > 0.$$
$$6k^2 - 3k^2 -6k - 3 = 3k^2 -6k - 3 = 3 (k^2 - 2k - 1) > 3 (k^2 - 2k - 3) = 3(k-3)(k+1).$$
Since we have assumed that $k \ge 8$...