Prove that $2 \cdot 3^n>3 n^2+6$ for natural number $n>2$

Question

Prove that $2 \cdot 3^n>3 n^2+6$ for natural number $n>2$.

I tried putting different values of $n$ and saw as $n$ increases RHS is indeed greater than RHS, and also their difference increases too. I know exponential functions increases more rapidly, but how do I prove this rigorously using elementary techniques only (like basic inequality like AM-GM, cauchy schartwz etc and not calculus).

By the way there exist N in N such that for all n>N this holds — , Nov 28 '22 at 21:04
@SafalDasBiswas I'm sorry I forgot to mention for all natural n>2 — Diff_Equations, Nov 28 '22 at 21:10
See this answer for the two generic approaches to proof by induction. — user2661923, Nov 28 '22 at 22:45

score 1 · Accepted Answer · answered Nov 28 '22 at 21:14

If $n\in\mathbb N$, then the given inequality holds for only $n≥3.$

The following proof uses the induction.

If $n=3$, then the inequality is correct.

Suppose that, the statemenet is also correct for $n=k$. Thus we have:

$$2\cdot 3^{k}-3k^2-6=a>0$$

Since $2\cdot 3^{k+1}>3(a+3k^2+6)$, then for $n=k+1$, we obtain:

$$\begin{align}&2\cdot 3^{k+1}-3(k+1)^2-6\\ >&~3(a+3k^2+6)-3(k+1)^2\\ =&~2k^2-2k+a+5\\ >&~2\left(k-\frac 12\right)^2+\frac 92\\ >&~0.\end{align}$$

By induction, the proof is completed.

Prove that $2 \cdot 3^n>3 n^2+6$ for natural number $n>2$

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