> [!NOTE] > For all $n\in \mathbb{Z}^+,$ $3^{n}\geq n^{3}.\tag{1}$ ###### Proof Suppose $(1)$ holds true for $n=k.$ Then $3^{k+1}= 3 \cdot 3^k \geq 3 k^3.$We want $3k^3 \geq (k+1)^3$ which is equivalent to $2k^3 \geq 3k^2+ 3k+ 1$which is true for $n \geq 3$ since $n^3 \geq 3n^2$ and $n^3 \geq 3n+1.$ Now $3^3=27\geq 27=3^3$ thus $(1)$ holds true for $n=3$ and all $n \geq 3$ by mathematical induction. We have also - $3^1=3\geq 1= 1^3.$ - $3^2 =9 \geq 8 = 2^3$ that is, $(1)$ hold true for $n=1,2$ also.