Prove by induction that $(k + 2)^{k + 1} \leq (k+1)^{k +2}$

Question

Prove by induction that $$ (k + 2)^{k + 1} \leq (k+1)^{k +2}$$ for $ k > 3 .$

I have been trying to solve this, but I am not getting the sufficient insight.

For example, $(k + 2)^{k + 1} = (k +2)^k (k +2) , (k+1)^{k +2}= (k+1)^k(k +1)^2.$

$(k +2) < (k +1)^2 $ but $(k+1)^k < (k +2)^k$ so what I want would clearly not be immediate from using something like If $ 0 < a < b, 0<c<d $ then $0 < ac < bd $. THe formula is valid for n = 4, So if it is valid for $n = k$ I would have to use

$ (k + 2)^{k + 1} \leq (k+1)^{k +2} $ somewhere in order to get that $ (k + 3)^{k + 2} \leq (k+2)^{k +3} $ is also valid. This seems tricky.

I also tried expanding $(k +2)^k $ using the binomial formula and multiplying this by $(k + 2)$, and I expanded $(k+1)^k$ and multiplied it by $(k + 1)^2 $ term by term. I tried to compare these sums, but it also gets tricky. I would appreaciate a hint for this problem, thanks.

Answer 1

For $k=4$ it's true.

Let $(k+2)^{k+1}\leq(k+1)^{k+2}.$ Thus, $$((k+2)^2)^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1}$$ or $$((k+1)(k+3)+1)^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1},$$ which gives $$((k+1)(k+3))^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1}$$ or $$(k+3)^{k+1}\leq(k+1)(k+2)^{k+1}.$$ Thus, $$(k+3)^{k+2}\leq(k+3)(k+1)(k+2)^{k+1}=$$ $$=(k^2+4k+3)(k+2)^{k+1}\leq(k+2)^2(k+2)^{k+1}=(k+2)^{k+3}$$ and we are done!

Answer 2

Try taking log of both sides and prove $\frac{\log x}x$ is decreasing.

Or by induction try to show $(\frac{k+1}k)^k\leq k$:

$$(1+1/k)^k\leq \sum_{i=0}^k \binom ki k^{-i}<\sum_{i=0}^k 1=k+1$$

Answer 3

A simpler (and a little stronger) statement is: $$\forall n\ge3\quad(n+1)^n<n^{n+1}.$$ We first check that $(3+1)^3=64<81=3^{3+1}.$ Then, for the induction step, it is sufficient to prove (for all $n\ge3$) that $$\frac{(n+2)^{n+1}}{(n+1)^{n+2}}\le\frac{(n+1)^n}{n^{n+1}},$$ i.e. $$(n(n+2))^{n+1}\le(n+1)^{2n+2},$$ i.e. $$n(n+2)\le(n+1)^2,$$ which is obvious.