Let us prove that 3n3+2≤n4 for all integers n≥4.
If n=4, then 3⋅43+2=194<256=44.
Suppose that 3k3+2≤k4, where k≥4, and prove the statement for k+1.
Since k≥4, we get that 2k≥8, and hence 2k−1≥7. Then (2k−1)2≥49, that is 4k2−4k+1≥49. Therefore, 4k2≥4k+48.
It follows that
9k2+9k+1=(3k2+5k)+(6k2+4k+1)=(3k+5)k+(6k2+4k+1)≤(4k+48)k+(6k2+4k+1)≤4k2k+(6k2+4k+1)=4k3+6k2+4k+1.
Consequently,
3(k+1)3+2=3k3+9k2+9k+3=(3k3+2)+(9k2+9k+1)≤k4+4k3+6k2+4k+1=(k+1)4.
According to mathematical induction principle, the statement 3n3+2≤n4 is true for all integers n≥4.
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