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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: $n=8$



Therefore true for $n=8$

Step $2$: Ass $n=k$


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


LHS=$2^k \cdot 2$

$>2 \cdot (3k^2)$


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


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
up vote 2 down vote accepted

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

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 $k \ge 8$...

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

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I doubt the OP is still stuck there after 18 months. – G. Sassatelli Sep 2 '15 at 9:39
What do you mean by "OP"? Sorry, I'm not so familiar with English and this site. – Kuh Crow Sep 4 '15 at 9:54
On math.SE, "Original Poster". – G. Sassatelli Sep 4 '15 at 10:57
Thanks. I agree with you! – Kuh Crow Sep 4 '15 at 11:13

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