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Prove that $m^3 \leq 2^m$ for $m \geq 10$

This is my shot

Proof

We prove this by induction.

Basic step: For $n=10$ is $10^3 \leq 2^{10} = 1000 \leq 1024$ and that is correct.

Induction step: Assume that $k \geq 10$ and that the statement is true for each $10 \leq j \leq k$. We need to proof that it is also true for $k+1$.

So $$(k+1)^3 \leq 2^{k+1} \Rightarrow k^3 + 3k^2 + 3k + 1 \leq 2^k + 2^k \underbrace{\Rightarrow }_{\text{induction hypothesis}} 3k^2 + 3k + 1 \leq 2^k$$

Now I don't understand how I can prove the rest.

NimaJan
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3 Answers3

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Your second equivalence is wrong. It has to be: $$k^3 + 3k^2 + 3k + 1 \leq 2^k + 2^k \impliedby k^3 \leq 2^k \land 3k^2 + 3k + 1 \leq 2^k$$

Now $k^3 \leq 2^k$ by the induction hypothesis.

For the last inequality, it is much easier to prove $3k^2+3k+1\leq k^3$, because $k^3 \leq 2^k$ by the induction hypothesis.

Now you just have to prove $-k^3+3k^2+3k+1\leq 0$ (for $k \geq 10$) which can be done by simple function analysis.

Victor Neverland
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$$(k+1)^3 = k^3+3k^2+3k+1 \leq k^3+3k^2+3k^2+4k^2 = k^3+10k^2$$

From here, we use that $k\geq 10$

$$k^3+10k^2\leq k^3+k^3$$

Next, we use the induction hypothesis

$$k^3+k^3\leq 2^k+2^k=2^{k+1}$$

JMoravitz
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    How did you come up to the idea to make this step $k^3+3k^2+3k+1 \leq k^3+3k^2+3k^2+4k^2$ – NimaJan Jan 06 '20 at 19:13
  • @NimaKimi From experience I knew that I wanted to arrive at $k^3+k^3$ and I figured that the $k^2$ and lower terms could all be combined into something less than $k^3$, despite their coefficients. I could have gotten away with much smaller... saying instead $3k^2+3k+1\leq 3k^2+k^2+k^2=5k^2$ noting that $k\geq 10\geq 3$ rather than $10k^2$ or something similar. With a proof like this you can be extremely loose with your inequalities and pick things that work and make sense to you. – JMoravitz Jan 06 '20 at 19:21
  • As a general strategy, when trying to show one polynomial is less than or equal to another polynomial of higher degree, I like to try increasing the degree of each term on the left to the highest degree and combine the terms into a single term which is generally much easier to work with. – JMoravitz Jan 06 '20 at 19:25
  • your explanation made it a lot more clear! Thanks. – NimaJan Jan 06 '20 at 19:32
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You have to prove $$m^3\le 2^m\implies (m+1)^3\le 2^{m+1}.$$

Taking the ratio,

$$\left(\frac{m+1}m\right)^3\le2\iff \frac{m+1}m\le\sqrt[3]2.$$

This inequality holds as soon as $m=4$.