Prove the inequality $(n+1)^4 < 4n^4$ for $n\geq 3$ by induction

Question

The inequality I'm concerned with is $(n+1)^4 < 4n^4,\ n\geq 3$.

I'm not sure how induction is supposed to work here. If I assume $(k+1)^4<4k^4$, I cannot see how this helps show $(k+2)^4<4(k+1)^4$. If I expand $(k+2)^4$ to get a polynomial in $k$, the induction hypothesis isn't useful (I can replace $k$ with $(k+1)$ only in the wrong direction! that is, I start decreasing from $(k+2)^4$ rather than increasing.)

If I expand to get a polynomial in $(k+1)$, the hypothesis is only useful in rewriting in terms of $k$.

Answer 1

note This proof I started aimed to use induction, but turned into a direct proof as I never used the induction hypothesis.

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Base case: $n=3$, we have $(3+1)^4 = 256 < 324 = 4\cdot 81 = 4\cdot (3)^4$

Assume it is true for our induction hypothesis that $(n+1)^4 < 4n^4$ for some $n\geq 3$

We continue to try to show it is true for $n+1$ as well.

$((n+1)+1)^4 = (n+1)^4 + 4(n+1)^3 + 6(n+1)^2 + 4(n+1) + 1$

as $n\geq 3$ and $n+1\geq 4$

$\leq (n+1)^4 + (n+1)(n+1)^3 + (n+1)^2(n+1)^2 + 4(n+1) + (n+1)$

$=3(n+1)^4 + 5(n+1)$

$\leq 4(n+1)^4$

thus proving the claim. (I point out again, this didn't require the use of the induction hypothesis, only the binomial expansion of the term on the left, and the fact that $n\geq 3$ and by replacing several constants by something larger (namely powers of $(n+1)$) and adding an extra $n$)

Answer 2

We need to prove that $$\frac{k+1}{k}<\sqrt 2\implies \frac{k+2}{k+1}<\sqrt 2,$$ Equivalent to saying that $$1+\frac 1k < \sqrt 2\implies 1+\frac 1{k+1}<\sqrt 2.$$ The last one is obvious as $\dfrac 1k>\dfrac 1{k+1}$.