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I am trying to find the point where the inequality: $6n^2 + 20n \leq n^3$ becomes true. How can I reduce it to see where it becomes true?

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    Is $n$ natural? – 2017-02-14
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    @S.C.B. What do you mean? – 2017-02-14
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    Meaning, is $n$ a real number or an integer? ($n$ is usually used to denote a natural number, i.e. a positive integer possibly including 0) – 2017-02-14
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    Yes n is a natural number – 2017-02-14
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    @user2896120 Then the equality condition is never met. – 2017-02-14

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Note that your inequality is equivlent to $$n^3 \ge 6n^2+20n \iff n^3-6n^2-20n \ge 0$$ Now factor this as so

$$n(n^2-6n-20) \ge 0 \iff n(n-3+\sqrt{29})(n-3-\sqrt{29})\ge 0$$ The equality occurs when $n=0, -\sqrt{29}+3 , 3+\sqrt{29}$. However, as $n$ is a natural, the equality never occurs. The inequality is true when $$n \ge3+ \sqrt{29}$$

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    So it would be safe to assume that the inequality is true when n > 8 correct? – 2017-02-14
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    @user2896120 Yes. – 2017-02-14