TITLE: How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$
QUESTION [11 upvotes]: show  that
  $$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$

it's well know that 
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I want see don't use $e=2.718$,because a most middle stundent don't know this value.
before I have use this well know
$$(1+\dfrac{1}{2n+1})(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e\cdot\dfrac{33}{34}\approx 2.638<\dfrac{8}{3}$$ to solve this, But Now we don't use $e=2.718$.
to prove this inequality by hand

REPLY [3 votes]: This can be done by hand as a fun little exercise in hexadecimal arithmetic, with some clever up-rounding to keep the calculations from getting too tiresome.  Writing everything (including the exponents) in base $16$, with digits $0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F$, the inequality we need to prove can be rewritten as
$$3\cdot11^{10}\lt8\cdot10^{10}$$
Now $11^2=121$ and $121^2=14641$, whether you do the calculations base $10$ or base $16$ (there are no carries in either case).  To go further, it helps to use the inequality
$$14641^2\lt14700\cdot14600$$
If this inequality doesn't strike you as obvious (and it shouldn't, really, since we're working in an unfamiliar base), note that
$$14641^2=(14700-BF)(14600+41)=14700\cdot14600-(146\cdot BF-147\cdot41)100-BF\cdot41$$
and
$$147\cdot41\lt200\cdot50=A000\lt B000=100\cdot B0\lt146\cdot BF$$
Continuing, we have
$$11^8=14641^2\lt14700\cdot14600=(147\cdot146)\cdot10^4=1A05A\cdot10^4\lt1A1\cdot10^6$$
and thus
$$11^{10}\lt1A1^2\cdot10^C=2A741\cdot10^C\lt2A800\cdot10^C=2A8\cdot10^E$$
so, finally,
$$3\cdot11^{10}\lt3\cdot2A8\cdot10^E=7F8\cdot10^E\lt800\cdot10^E=8\cdot10^{10}$$
as desired.
Please note, I did all the three-digit multiplications here literally by hand, on paper, so I hope someone will take the time to check my arithmetic and correct it as necessary.  The crucial base-$16$ calculations that aren't eyeballable are
$$\begin{align}
121^2&=14641\\
14641&=14700-BF\\
147\cdot146&=1A05A\\
1A1^2&=2A741\\
3\cdot2A8&=7F8\\
\end{align}$$