Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
 A: Suppose $2k^3<3^k$. Then
\begin{align}
2(k+1)^3&=2(k^3+3k^2+3k+1)\\
&=2k^3+6k^2+6k+2\\
&<3^k+6k^2+6k+2\\
&< 3^k+k^3+k^2+k\\
&<3^k+4k^3\\
&<3^k+2\cdot3^k\\
&=3^{k+1}
\end{align}
Note that we use in the middle that $6\le k$.
A: Note that $f(k) = \frac{k}{k+1}$ is a positive, strictly increasing function for positive $k$ (since $\frac{k}{k+1} = 1 - \frac{1}{k+1}$ and $\frac{1}{k+1}$ is strictly decreasing). Thus, for $k \ge 6$, you have
$$\begin{equation}\begin{aligned}
f^3(k) & = \left(\frac{k}{k+1}\right)^3 \\
& \ge \left(\frac{6}{7}\right)^3 \\
& = \frac{216}{343} \\
& \gt \frac{1}{3}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Using this, you get with the induction part that
$$\begin{equation}\begin{aligned}
3^{k+1} & = 3(3^{k}) \\
& > 3(2k^3) \\
& = 2(3)\left(\frac{k}{k+1}\right)^3(k+1)^3 \\
& \gt 2(3)\left(\frac{1}{3}\right)(k+1)^3 \\
& = 2(k + 1)^3
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, this shows that if $P(k)$ is true, then so is $P(k+1)$.
A: $P(k)$ is the statement $ 2n^3<3^n $.  Do not write "$P(k)=.....$"; $P(k)$ is not a mathematical value.
If we assume that then we have $2n^3 < 3^n$
So $2(n+1)^3 = 2n^3 + 6n^2 + 6n + 1$  And we have $2n^3 < 3^n$ so
$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + 6n^2 + 6n + 1$
And $n\ge 6$ so $6n^2 \le n*n^2 =n^3$ and $6n+1 < 6n+n < 6n*n=6n^2 < n*n^3 < n^3$.
So 
$2(n+1)^3 =2n^3 + 6n^2 + 6n + 1< 3^n + (6n^2) + (6n + 1)$
$< 3^n + n^3 + n^3 = 3^n + 2n^3 \le 3^n + 3^n < 3^n + 3^n + 3^n$
$< 3*3^n = 3^{n+1}$.
So $2(n+1)^3 < 3^{n+1}$ and so the statement $P(k+1)$ is true.
A: We have to prove that for $n \geq 6$ it is
$$
3^{h + 1}  > 2\left( {h + 1} \right)^3 
$$
This is equivalent to 
$$
3^h  + 3^h  + 3^h  > 2h^3  + 6h^2  + 6h + 2
$$
We have that


*

*$3^h>2h^3$ by the inductive hypothesis.

*$3^h>2h^3>h^3\geq 6h^2$  by the inductive hypothesis and since $h \geq 6$.

*$3^h>2h^3>h^3>6h+2$ since
$$
h^3  - 6h = h\left( {h^2  - 6} \right) > 2
$$
as long as $h^2-6>0$. Namely, in this case it is $h^2-6h \geq 1$ and $h(h^2-6h) \geq 6$
A: $$P(k+1)=2(k+1)^3<\left(\dfrac{k+1}k\right)^33^k$$
So, it is sufficient to establish $\left(\dfrac{k+1}k\right)^3<3$
which is true if $\dfrac1k<\sqrt[3]3-1\iff k>\dfrac1{\sqrt[3]3-1}$ 
Now  $3>\dfrac1{\sqrt[3]3-1}\iff\dfrac13+1<\sqrt[3]3\iff\dfrac{64}{27}<3\iff64<81$ which is true
So, $\left(\dfrac{k+1}k\right)^3<3$ if $k\ge3$
Now establish  $P(6)$
