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Sunday, March 29, 2020

2020/011) Prove that 4^{79}< 2^{100} + 3^{100} < 4^{80}

We need to prove 2 inequalities
4^{79}< 2^{100} + 3^{100} and 2^{100} + 3^{100} < 4^{80}

For

4^{79}< 2^{100} + 3^{100}

If we prove that  4^{79} < 3^{100} we are through

We have \frac{4^{80}}{3^{100}} = (\frac{4^4}{3^5})^{20} = (\frac{256}{243})^{20}\cdots(1)

Now \frac{256}{243} = 1+\frac{13}{243} < 1 + \frac{13}{13 *18} = 1 + \frac{1}{18} = \frac{19}{18}\cdots(2)

Hence (\frac{256}{243})^{20} < (\frac{19}{18})^{20} = (\frac{361}{324})^{10}=  (1 + \frac{37}{324})^{10}\cdots(2)

We have
\frac{37}{324} < \frac{37}{37 * 8} < \frac{1}{8}

From (2) and above
(\frac{256}{243})^{20} < (1 + \frac{1}{8})^{10} = (\frac{9}{8})^{10} = (\frac{81}{64})^5 < (\frac{81}{63})^5 = (\frac{9}{7})^5 = \frac{59049}{16807} < 4

From (1) and above \frac{4^{80}}{3^{100}} < 4 and hence 4^{79} < 3^{100}

Hence 4^{79} < + 2^{100} + 3^{100}

For proving the 2^{nd} part we have

  (\frac{256}{243})^20 = (1+  \frac{13}{243})^{20} > 1+  \frac{13}{243} * 20  > 1+ \frac{260}{243} > 2 by bionomial expansion and deleting positive terms

Form above and (1) we have

\frac{4^{80}}{3^{100}} > 2 or 4^{80} > 2 * 3^{100} > 3^{100} + 2^{100}

Proved

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