Prove that $2\left(\frac{6}{e}\right)^e>17$

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?

• Why don't you take $\ln$ of both sides and deconstruct the left hand side using log laws? Commented Aug 21, 2017 at 14:53
• Maybe "easy to notice" means use a calculator? Commented Aug 21, 2017 at 14:57
• Please tell us which textbook it is. Commented Aug 21, 2017 at 16:27

$$2\left(\frac{6}{e}\right)^e>17.$$ $$e=2.718$$ 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
$\ln2=0.693$, $\ln6=1.791$, $\ln17=2.833$ $$0.693+2.718\times(1.791-1)>2.833$$ $$2.842>2.833$$