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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