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## Here's the Solution to this Question

Let us prove that $3n^3+2≤n^4$ for all integers $n\ge4.$

If $n=4,$ then $3\cdot 4^3+2=194<256=4^4.$

Suppose that $3k^3+2≤k^4,$ where $k\ge 4,$ and prove the statement for $k+1.$

Since $k\ge 4,$ we get that $2k\ge 8,$ and hence $2k-1\ge 7.$ Then $(2k-1)^2\ge 49,$ that is $4k^2-4k+1\ge 49.$ Therefore, $4k^2\ge4k+48.$

It follows that

$9k^2+9k+1=(3k^2+5k)+(6k^2+4k+1) =(3k+5)k+(6k^2+4k+1)\\ \le (4k+48)k+(6k^2+4k+1)\le 4k^2k+(6k^2+4k+1)=4k^3+6k^2+4k+1.$

Consequently,

$3(k+1)^3+2=3k^3+9k^2+9k+3=(3k^3+2)+(9k^2+9k+1)\\ \le k^4+4k^3+6k^2+4k+1=(k+1)^4.$

According to mathematical induction principle, the statement $3n^3+2≤n^4$ is true for all integers $n\ge4.$

Let us prove that $3n^3+2≤n^4$ for all integers $n\ge4.$

If $n=4,$ then $3\cdot 4^3+2=194<256=4^4.$

Suppose that $3k^3+2≤k^4,$ where $k\ge 4,$ and prove the statement for $k+1.$

Since $k\ge 4,$ we get that $2k\ge 8,$ and hence $2k-1\ge 7.$ Then $(2k-1)^2\ge 49,$ that is $4k^2-4k+1\ge 49.$ Therefore, $4k^2\ge4k+48.$

It follows that

$9k^2+9k+1=(3k^2+5k)+(6k^2+4k+1) =(3k+5)k+(6k^2+4k+1)\\ \le (4k+48)k+(6k^2+4k+1)\le 4k^2k+(6k^2+4k+1)=4k^3+6k^2+4k+1.$

Consequently,

$3(k+1)^3+2=3k^3+9k^2+9k+3=(3k^3+2)+(9k^2+9k+1)\\ \le k^4+4k^3+6k^2+4k+1=(k+1)^4.$

According to mathematical induction principle, the statement $3n^3+2≤n^4$ is true for all integers $n\ge4.$

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