<s>[INST] $n^3+n<3^n$ for $n \geq4$ by induction. To prove that $n^3+n<3^n$ for $n \geq4$ by induction.
I have proved the fact but it became very long and I have to use two more induction proof within the proof. 
Can someone give a better solution by induction?
Thank You.


 [/INST] $n^3 + n < 3^n \tag 1$
certainly holds for $n = 4$; so this will be the base case; assuming (1) binds for some $n$, then
$3(n^3 + n) < 3^{n + 1}; \tag 2$
now,
$(n + 1)^3 + (n + 1) = n^3 + 3n^2 + 3n + 1 + n + 1 = n^3 + 3n^2 + 4n + 2, \tag 3$
and
$3(n^3 + n) = 3n^3 + 3n, \tag 4$
so what we need is
$n^3 + 3n^2 + 4n + 2 \le 3n^3 + 3n, \; n \ge 4; \tag 5$
this is true if
$3n^3 + 3n - n^3 - 3n^2 - 4n - 2 = 2n^3 - 3n^2 -n - 2 \ge 0, \; n \ge 4; \tag 6$
now, as our colleague xbh has observed,
$2n^3 - 3n^2 - n - 2 = (n - 1)(2n^2 - n - 2) - 4; \tag 7$
we have
$n \ge 4 \Longrightarrow n - 1 \ge 3; \tag 8$
also we have
$n \ge 4 \Longrightarrow 2n^2 - n - 2 = n^2 +  n(n - 1) - 2 \ge  16 + 4 \cdot 3 - 2 = 26; \tag 9$
thus,
$n \ge 4 \Longrightarrow 2n^3 - 3n^2 - n - 2 \ge 3 \cdot 26 - 4 = 74; \tag{10}$
we see that (6) binds, thus does (5), hence the induction is complete and (1) holds for all $n \ge 4$.
</s>