if-a-gt-b-gt-0-prove-that-b-lt-ae-x-be-x-e-x-e-x-lt-a-

Question Number 34365 by mondodotto@gmail.com last updated on 05/May/18

if a > b > 0 prove that

b < \frac{{ae}^{x} + {be}^{- x}}{e^{x} + e^{- x}} < a

Answered by MJS last updated on 05/May/18

b e^{x} + b e^{- x} < a e^{x} + b e^{- x} \Rightarrow

\Rightarrow b e^{x} < a e^{x}

a e^{x} + b e^{- x} < a e^{x} + a e^{- x} \Rightarrow

\Rightarrow b e^{- x} < a e^{- x}

both are true

Facebook Tweet Pin