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

$$\mathbf{if}\mathbf{a}>\mathbf{b}>0\mathbf{prove}\mathbf{that}$$ $$\mathbf{b}<\frac{{\mathbf{ae}}^{\mathit{x}}+{\mathbf{be}}^{-\mathit{x}}}{{\mathbf{e}}^{\mathit{x}}+{\mathbf{e}}^{-\mathit{x}}}<\mathbf{a}$$

Answered by MJS last updated on 05/May/18

$$b{\mathrm{e}}^x+b{\mathrm{e}}^{-x}<a{\mathrm{e}}^x+b{\mathrm{e}}^{-x}\Rightarrow$$ $$\Rightarrow b{\mathrm{e}}^x<a{\mathrm{e}}^x$$ $$a{\mathrm{e}}^x+b{\mathrm{e}}^{-x}<a{\mathrm{e}}^x+a{\mathrm{e}}^{-x}\Rightarrow$$ $$\Rightarrow b{\mathrm{e}}^{-x}<a{\mathrm{e}}^{-x}$$ $$\mathrm{both}\mathrm{are}\mathrm{true}$$

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