Question: > 
> 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 $3n^2 + 3n +2$.

Answer: 
Hint:
$$ 3n^2+3n+2<n^3+n<3^n. $$
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.