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 $3n^2 + 3n +2$. 
 A: Hint:
$$ 3n^2+3n+2<n^3+n<3^n. $$
A: Because for $n\geq4$ we have:
$$3^n=(1+2)^n\geq1+2n+\frac{n(n-1)}{2}\cdot2^2+\frac{n(n-1)(n-2)}{6}\cdot2^3+$$
$$+\frac{n(n-1)(n-2)(n-3)}{24}\cdot2^4>n^3+n.$$
The last inequality is true because it's
$$3+6n+6n^2-6n+4n^3-12n^2+8n+2n^4-12n^3+22n^2-12n>3n^3+3n$$ or
$$2n^4-11n^3+16n^2-7n+3>0$$ or
$$2n^4-8n^3-3n^3+12n^2+4n^2-16n+9n+3>0$$ or
$$(n-4)(2n^3-3n^2+4n)+9n+3>0,$$
which is obvious for $n\geq4$.
A: You proved correctly that $(n+1)^3+(n+1)<3^n+3n^2+3n+2$. So, prove now that$$(\forall n\geqslant4):3n^2+3n+2\leqslant2\times3^n.$$This is easy to do using induction again.
A: 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.
