Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$ 
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$

My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then, 
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3n^2 + 3n +2$. 

By the binomial theorem, $3^n = (1+2)^n \ge \binom{n}{1}+ \binom{n}{4}2^4$ for $n \ge 4$.
So, it's enough to prove that $\binom{n}{1}+ \binom{n}{4}2^4 > n^3+n$ (*).
This simplifies to $2 n^3 - 15 n^2 + 22 n - 12 > 0$.
The largest critical point of $2 x^3 - 15 x^2 + 22 x - 12$ is at $x \approx 4.1$. Also (*) is true for $n=6$. Therefore, (*) is true for all $n \ge 6$.
The cases $n=4$ and $n=5$ are easily handled separately.
