Find all natural numbers n such that $7n^3 < 5^n$ Find all natural numbers $n$ such that $7n^3 < 5^n$.
I drew a graph which showed that $n \geq 4$
wolfram
How can I prove that? I guess I need to use induction with the base case $n=4$?
But I am stuck because the induction hypothesis uses a $\geq$ sign so I can not substitute...
 A: Base Case: For $n=4$, we have $7(4)^3 = 448 < 625 = 5^4$, which works.
Induction Hypothesis: Assume that the claim holds true for all $n \in \{4,\ldots,k\}$, where $k>3$.
It remains to prove the inequality true for $n=k+1$. Indeed, observe that:
\begin{align*}
7(k+1)^3 &= 7(k^3 + 3k^2 + 3k + 1) \\
&< 7(k^3 + 3k^2 + 9k + 27) \\
&= 7(k^3 + (3)k^2 + (3)^2k + (3)^3) \\
&< 7(k^3 + (k)k^2 + (k)^2k + (k)^3) & \text{since }3 < k\\
&= 7(4k^3) \\
&= 4(7k^3) \\
&< 5(7k^3) \\
&< 5(5^k) & \text{by the induction hypothesis}\\
&= 5^{k+1}\\
\end{align*}
as desired. This completes the induction.
A: First, check for $n = 4$:
$$7n^3 = 448 < 625 = 5^n.$$
Now, assume that it is true for all numbers strictly less than $n$. Let us check for $n$:
\begin{align*}
7n^3 &= 7((n-1)+1)^3 = 7(n-1)^3 + 7 \cdot 3 \cdot (n-1)^2 + 7 \cdot 3 \cdot (n-1) + 7. %\\
%&< 4 \cdot 7 (n-1)^3 < 4 \cdot 5^{n-1} < 5 \cdot 5^{n-1} = 5^n.
\end{align*}
Here, use that $3(n-1)^k < (n-1)^3$ for $k \in \{1,2\}$, which comes from the fact that $n > 4$, so $n - 1 > 3$. Also, $7 < (n-1)^3$. If you get stuck, ask.
