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I need to prove that

$$|2n^5 − n^3 + 2000| ≥ 2n^5 − 2001n^4$$

For $n$ in the natural numbers.

I understand to do this I need to use a Triangle Inequality. So far I have,

$$|2n^5-n^3+2000|≥ 2n^5 - 2001n^4$$

$$\implies |2n^5|-|n^3+2000| ≥ 2n^5 - 2001n^4$$

From here, I am not quite sure where to go.

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    $\begingroup$ You posted a very similar question about 40 minutes ago. Can you use the answers from that question to answer this one? $\endgroup$ – Χpẘ Apr 21 '17 at 1:36
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$|2n^5-n^3+2000| = |2n^5-(n^3-2000)| \ge |2n^5| - |n^3-2000|=2n^5-|n^3-2000|$.

Thus it remains to show: $-|n^3-2000| \ge -2001n^4$ or: $2001n^4 \ge$

$|n^3-2000|$ ,but this is clear since $|n^3-2000| \le |n^3| + |-2000|= n^3 +$

$2000 \le n^4 + 2000 \le n^4 + 2000n^4 = 2001n^4$.

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Since $2n^5-n^3=n^3(2n^2-1)>0$ the absolute value on the left side does not work and the inequality becomes as follows $$2n^5-n^3+2000\geq 2n^5-2001n^4\Leftrightarrow 2001n^4-n^3+2000\geq 0.$$ This is evidently true because $2001n^4-n^3+2000=n^3(2001n-1)+2000\geq 2000+2000>0.$

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