prove-that-e-2-3-gt-13-2-6-2-

Question Number 147183 by mathdanisur last updated on 18/Jul/21

p r o v e t h a t

e^{\sqrt{2} + \sqrt{3}} > \frac{13 + 2 \sqrt{6}}{2}

Answered by mindispower last updated on 18/Jul/21

\Leftrightarrow e^{\sqrt{2} + \sqrt{3}} > 1 + 3 + \frac{1}{2} (5 + 2 \sqrt{6}) (E)

w e s h o w \sqrt{2} + \sqrt{3} > 3 \dots (1)

\Rightarrow 5 + 2 \sqrt{6} > 9

\Rightarrow \sqrt{6} > 2 . . t r u e

i f e^{\sqrt{2} + \sqrt{3}} > 1 + \sqrt{2} + \sqrt{3} + \frac{1}{2} (\sqrt{5} + 2 \sqrt{6}) T h i s

5 + 2 \sqrt{6} = {(\sqrt{2} + \sqrt{3})}^{2}

\Rightarrow (E) T r u e

l e t t = \sqrt{2} + \sqrt{3}

f (x) = e^{x} - 1 - x - \frac{x^{2}}{2}

x > 0, j u s t u s i n g e^{x} = \underset{k = 0}{\sum^{\infty}} \frac{x^{k}}{k!}

\Rightarrow f (x) = \underset{n = 3}{\sum^{\infty}} \frac{x^{n}}{n!} > 0

\Rightarrow f (t) > 0 \Leftrightarrow e^{t} > 1 + t + \frac{t^{2}}{2}, t = \sqrt{2} + \sqrt{3} > 3

\Rightarrow e^{\sqrt{2} + \sqrt{3}} > 1 + 3 + \frac{1}{2} (5 + 2 \sqrt{6}) = \frac{13 + 2 \sqrt{6}}{2}

Commented by mathdanisur last updated on 19/Jul/21

t h a n k s S e r c o o l

Commented by mindispower last updated on 19/Jul/21

w i t h e p l e a s u r