Showing $\sqrt{\frac{e}{2}}\cdot\frac{e}{\pi}\left(\frac{e}{2}-\frac{1}{e}\right)<1$ without a calculator

Solution 1:

A piece of sheer numerical good luck makes the calculation fairly painless.

I'll take this inequality to be well known: $$ \pi > \frac{333}{106}. $$ (I don't know a simple proof. For some that aren't simple, see Is there an integral that proves $\pi > 333/106$?.)

From $e$ Continued Fraction --- from Wolfram MathWorld, we have: $$ e < \frac{193}{71}. $$

Therefore, it is enough to prove: $$ \sqrt{\frac{193}{2\cdot71}} < \frac{333}{106}\left(\frac{193^2}{2\cdot71^2} - 1\right)^{-1}. $$ But \begin{multline*} 193^2 - 2\cdot71^2 = (200 - 7)^2 - 2(70 + 1)^2 = (40000 - 2800 + 49) - 2(4900 + 140 + 1) \\ = 37249 - 10082 = 27167 = 7\cdot3881, \end{multline*} and the right hand side of the required inequality simplifies to: $$ \frac{2\cdot3\cdot111\cdot71^2}{2\cdot7\cdot53\cdot3881} = \frac{426\cdot7881}{371\cdot7762} > \frac{426\cdot7882}{371\cdot7763} = \frac{426\cdot1126}{371\cdot1109} = \frac{426}{371}\left(1 + \frac{17}{1109}\right), $$ where we have used the fact that if $y > x > 0$ then $\frac{y}{x} > \frac{y + 1}{x + 1}.$

Now comes the piece of luck: $$ \frac{1109}{17} = 65 + \frac4{17} < 65 + \frac14 = \frac{261}4, \text{ whence } 1 + \frac{17}{1109} > 1 + \frac4{261}, $$ whence the right hand side of the required inequality is greater than: $$ \frac{426\cdot265}{371\cdot261} = \frac{(6\cdot71)\cdot(5\cdot53)}{(7\cdot53)\cdot(3\cdot87)} = \frac{710}{609}. $$ We now only have to prove: $$ \frac{193}{2\cdot71} < \left(\frac{710}{609}\right)^2, \text{ i.e., }\ 200\cdot71^3 > 193\cdot609^2. $$ We have already calculated $71^2 = 5{,}041,$ so $71^3 = 352{,}870 + 5{,}041 = 357{,}911,$ and $200\cdot71^3 = 71{,}582{,}200.$ On the other hand, $609^2 = 360{,}000 + 10{,}800 + 81 = 370{,}881,$ therefore $193\cdot609^2 = (200 - 7)\cdot370{,}881 = 74{,}176{,}200 - 2{,}596{,}167 = 71{,}580{,}033.$ This proves the inequality.

Solution 2:

Hint:

By excess rational approximations of $e$ can be found from the usual Taylor expansion.

By default rational approximations of $\pi$ can be found from the Machin formula, among others.

Five exact decimals should be enough.

Now the following inequality can be established by hand (though this will take several hours):

$$\frac{\overline{e}(\overline{e}^2-2)^2}{\underline{\pi}^2}<8.$$

(With $\overline e=2.71829$ and $\underline\pi=3.14159$, you get $7.99888\dots<8$.)

Not glamorous, but effective.

Solution 3:

Using values rounded up to the fourth decimal,

$$e(e^2-2)<2.7183(2.7183^2-2)^2<2.7183\cdot5.3891^2<2.7183\cdot29.0424<78.9460$$

and down

$$78.9520=8\cdot9.8690<8\cdot3.1415^2<8\pi^2.$$

This takes four multiplies of $4$-decimal numbers and a little more.

(If using precomputed tables is allowed, this one supplies $\sqrt2,\pi,\pi^2,\sqrt e,e,e^2$: http://www.ebyte.it/library/educards/constants/MathConstants.html)