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Proof by induction involving inequalities completely escapes me. I've encountered the following problem:

For which non-negative integers n is $n^2 ≤ n!$? Prove your answer (by induction).

So, computing the first few values of n leads to the conjecture that $n^2 ≤ n!$ for 0,1, and values greater or equal to 4.

My base step is P(4) : $4^2 ≤ 4!$ which holds

My inductive hypothesis is P(k+1) for nonnegative integer values of k greater or equal to 4.

My inductive step needs to prove that $(k+1)^2\le(k+1)!$

I've encountered two different solutions for this problem but I can't figure out how they were arrived at. Take the following for instance: $$ (k+1)^2 = k^2 + 2k + 1 \le k! + 2k + 1 \le k! + k!+ k! \le 3 (k!)\le (k+1)k! = (k+1)!$$

The first piece of this solution makes sense: $(k+1)^2 = k^2+2k+2 \le k! + 2k + 2$ but the rest of it eludes me, specifically $\le k! + k!+ k! \le 3(k!)$... where is this coming from? How do I know that $3 (k!)\le (k+1)k! = (k+1)!$ is true?

Proof by induction for summations and equalities(?) seem fairly accessible to me. Inequalities just don't seem to register. Any insight would be appreciated.

David K
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nerraga
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2 Answers2

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If $m!\ge m^2,$

$$(m+1)!=(m+1)\cdot m!\ge(m+1)\cdot m^2$$ which needs to be $\ge(m+1)^2$

$$\iff m^2\ge m+1\iff m(m-1)\ge1$$ which holds true if $m\ge2$

lab bhattacharjee
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As you claim, $n^2 \leq n!$ holds for $n = 4$. You now are assuming that $k^2 \leq k!$ for some $k \geq 4$ as your inductive hypothesis.

Since $k \geq 4$, $k+1 \geq 5 > 3$. Multiplying by $k!$ yields $(k+1)k! \geq 3k!$.

As you note, by the inductive hypothesis, we already see that $k^2 \leq k!$.

Since $4 \leq k$, $2 \leq k \implies 2k \leq k^2 \leq k!$.

Also, $1 \leq k \implies k \leq k^2$ so we can see that $1 \leq k \leq k^2 \leq k!$.

C. Caruvana
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