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