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How would you prove that $8n^5 > (n+4)^5$ for $n \ge 8$ by induction?

I can't figure out any link between $P(n)$ and $P(n+1)$, Any hint?

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    $\begingroup$ It is possible to prove it by induction. $\endgroup$ – kingW3 Mar 19 '17 at 10:20
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It is same as establishing $$\left(1+\dfrac4n\right)^5<8$$

Now observe that $$\left(1+\dfrac4{m+1}\right)^5<\left(1+\dfrac4m\right)^5$$

or more generally $$\left(1+\dfrac4q\right)^5>\left(1+\dfrac4r\right)^5$$ for $q<r$

So it is sufficient to show $$\left(1+\dfrac48\right)^5<8$$

$$\iff3^5(=243)<8\cdot2^5(=256)$$

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Let $f(n)=(n+4)/n=1+4/n.$

  1. Show that for all $n\in \mathbb N$ we have $f(n+1)<f(n)$ and hence $f(n+1)^5<f(n)^5$.

  2. Use 1. and induction on $n$ to show that $n\geq 8\implies f(n)^5\leq f(8)^5.$

  3. Observe that for any $n$ we have $8n^5>(n+4)^5\iff 8>f(n)^5.$

    By 2. we have $n\geq 8\implies f(n)\leq f(8)=243/32<8.$

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