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Prove that $$87!<16! \left(52^{71}\right)$$ I do not how can i compare between the factorials or what the procedure to solve such questions?

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    $\begingroup$ $87!=16! \times 17 \times \dots \times 87$ $\endgroup$ – Naj Kamp Jan 27 '18 at 13:02
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Note that

$$87!<16! \left(52^{71}\right)\iff \frac{87!}{16!}<52^{71}\iff\binom{87}{71}<\frac{52^{71}}{71!}$$

which is true from $\ { n \choose k} \leq \frac{n^k}{k!}. ( \, 1-\frac {k}{2n} ) \,^{k-1} $ since

$$\binom{87}{71}\le \frac{87^{71}}{71!}\cdot\left(1-\frac{71}{2\cdot 87}\right)^{70}=\frac{52^{71}}{71!}\cdot \frac{87^{71}}{52^{71}}\cdot \frac{103^{71}}{174^{71}}\cdot\frac{174}{103}=\frac{174}{103}\cdot\left(\frac{87\cdot103}{52\cdot 174}\right)^{71}\cdot \frac{52^{71}}{71!}$$

$$=\frac{174}{103}\cdot \left(\frac{103}{104}\right)^{71}\cdot \frac{52^{71}}{71!}<0.851\cdot \frac{52^{71}}{71!}$$

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You know that $87!=16! \cdot 17 \cdot 18 \dots 87$, hence you want to show that

$$17 \cdot 18\cdot 19 \dots 87 < 52^{71}$$

Now notice that on the left hand side you have $71$ terms with $52$ being the middle one. Hence you can group that in pairs (first with last, second with second last etc. and finally leave $52$ alone). You can say the following:

$$(52-x)(52+x)=52^2-x^2<52^2$$

Hence the product in each pair is less than $52^2$

Do this for all $x$'s between $1$ and $35$ to get that the product is less than $52^{71}$ since you have $35$ pairs for which it holds and then you add the $52$ factor

QED

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AM-GM gives: $$\frac1{71}\sum_{i=1}^{71}(i+16)>\sqrt[71]{\prod_{i=1}^{71}(i+16)}$$ So: $$\frac1{71}\left(\frac{71(71+1)}{2}+71\cdot 16\right)>\sqrt[71]{\frac{87!}{16!}}$$ or: $$52^{71}>\frac{87!}{16!}\implies 16!(52^{71})>87!$$

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  • $\begingroup$ That is great solution ,thank you @Mastrem $\endgroup$ – Hussien Mohamed Jan 27 '18 at 14:42

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