Can anybody tell me whether or not my proof by induction is correct?

Question

Statement,

$P(n) : n^4<4^n , n≥5 $

$P(5)$ is true.

And Assuming $P(k) : k^4 < 4^k$ is true for $k \in \mathbb{N}$ (Quick Question : Do I also need to mention $k≥5$ in my inductive hypothesis?)

Coming back to the question, I need to prove $P(k+1) : (k+1)^4<4^{k+1}$ is true whenever $P(k)$ is true.

My thoughts :

$$ k^4<4^k \Rightarrow 4k^4 < 4^{k+1}$$ and if somehow I could prove that $$(k+1)^4 <4k^4 $$ then I'm done.

For that I tried a couple of stunts, for instance, bringing $4k^4$ on the left side and applying $a^2-b^2$, so the $a+b$ term can be divided as it's positive. So I'm finally left with $k^2 -2k-1 >0$ I did this in order to see if the resulting expression is actually greater than $0$ for the required values of $k$, so my assumption that $(k+1)^4< 4k^4$ would then be true. Now I get $k=\{3,4,5,...\}$ so $(k+1)^4< 4k^4$ is true for $k≥3, k \in \mathbb{N}$ and since my base case starts from $5$, therefore I'm done.

So finally $(k+1)< 4k^4< 4^{k+1}$ hence, $(k+1)< 4^{k+1}$ proved.

Idk, I feel this is correct. But somehow it doesn't feel like the right way of solving this. Can anybody show me an alternative way of proving $P(k) \Rightarrow P(k+1)$ ?

Answer 1

TP: $(k+1)^4<4k^4\Longleftrightarrow3k^4-4k^3-6k^2-4k-1>0$

$\Longleftrightarrow(k^4-4k^3)+(k^4-6k^2)+(\frac45k^4-4k)+(\frac15k^4-1)>0$

which is trivially true for all $k>5$

Answer 2

Your reasoning is sound.

For an alternative, you could have noted that $$ (k+1)^4= k^4 +4k^3 + 6k^2 + 4k +1,$$

and that $$4k^4 = k^4 + k^4 + k^4 + k^4.$$

Next, use the assumption that $k\geq 5$: $$k^4= k k^3 \gt 4k^3,$$ $$k^4= k^2 k^2 \gt 6k^2, $$ $$k^4= k^3 k \gt 4k,$$

Thus $$4k^4 \geq k^4 + 4k^3 + 6k^2 + 4k +3 \gt (k+1)^4,$$ which is what you wanted to prove.

$\square$