<s>[INST] 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
 [/INST] 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 $P(k+1)$.
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