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Question Number 34365 by mondodotto@gmail.com last updated on 05/May/18
if a>b>0 prove that  b<((ae^x +be^(−x) )/(e^x +e^(−x) ))<a
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}>\mathrm{0}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{b}}<\frac{\boldsymbol{\mathrm{ae}}^{\boldsymbol{{x}}} +\boldsymbol{\mathrm{be}}^{−\boldsymbol{{x}}} }{\boldsymbol{\mathrm{e}}^{\boldsymbol{{x}}} +\boldsymbol{\mathrm{e}}^{−\boldsymbol{{x}}} }<\boldsymbol{\mathrm{a}} \\ $$
Answered by MJS last updated on 05/May/18
be^x +be^(−x) <ae^x +be^(−x)  ⇒  ⇒ be^x <ae^x   ae^x +be^(−x) <ae^x +ae^(−x)  ⇒  ⇒ be^(−x) <ae^(−x)   both are true
$${b}\mathrm{e}^{{x}} +{b}\mathrm{e}^{−{x}} <{a}\mathrm{e}^{{x}} +{b}\mathrm{e}^{−{x}} \:\Rightarrow \\ $$$$\Rightarrow\:{b}\mathrm{e}^{{x}} <{a}\mathrm{e}^{{x}} \\ $$$${a}\mathrm{e}^{{x}} +{b}\mathrm{e}^{−{x}} <{a}\mathrm{e}^{{x}} +{a}\mathrm{e}^{−{x}} \:\Rightarrow \\ $$$$\Rightarrow\:{b}\mathrm{e}^{−{x}} <{a}\mathrm{e}^{−{x}} \\ $$$$\mathrm{both}\:\mathrm{are}\:\mathrm{true} \\ $$

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