TITLE: Prove that $2\left(\frac{6}{e}\right)^e>17$
QUESTION [8 upvotes]: I was doing some exercises in my calculus textbook and to finish one, I need to prove that
$$2\left(\frac{6}{e}\right)^e>17.$$
This is supposed to be simple (,,easy to notice'', the textbook says). However, I cannot figure it out withour a calculator (17.2 is about the exact value of the left hand side). My only idea was to use Bernoulli inequality:
$$2\left(\frac{6}{e}\right)^e=2\left(1+\left(\frac{6}{e}-1\right)\right)^e\ge 2\left(1+e\left(\frac{6}{e}-1\right)\right)=2(7-e)=14-2e$$
but this is to weak. Could anybody give me some hints?

REPLY [2 votes]: $$2\left(\frac{6}{e}\right)^e>17.$$
$$e=2.718$$
Taking $\log$ to the base $e$ or '$\ln$'
To prove that L.H.S>R.H.S
$$\ln2+e(\ln6-1)>\ln17$$
Please see the log table
$\ln2=0.693$, $\ln6=1.791$, $\ln17=2.833$
$$0.693+2.718\times(1.791-1)>2.833$$
$$2.842>2.833$$