<s>[INST] 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 $3n^2 + 3n +2$. 
 [/INST] 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 $n\geq4$.
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