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Question Number 128731 by TITA last updated on 09/Jan/21

$$\mathrm{for}\mathrm{a}>\mathrm{b}>0\mathrm{show}\mathrm{that}$$ $$\mathrm{b}<\frac{{\mathrm{ax}}^{\mathrm{x}}+{\mathrm{bx}}^{-\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}<\mathrm{a}$$

Commented byTITA last updated on 09/Jan/21

$$\mathrm{please}\mathrm{help}$$

Commented byMJS_new last updated on 09/Jan/21

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{true}\mathrm{if}\mathrm{we}\mathrm{are}\mathrm{free}\mathrm{to}\mathrm{choose}x\in {\mathbb{R}}^{+}$$ $$\mathrm{i}.\mathrm{e}.$$ $$b=1\wedge a=2$$ $$1<\frac{2x^x+x^{-x}}{{\mathrm{e}}^x+{\mathrm{e}}^{-x}}<2\iff 0<x<\approx .937325\vee \approx 1.83623<x<\approx 2.72045$$ $$\mathrm{not}\mathrm{true}\mathrm{for}\mathrm{all}\mathrm{other}\mathrm{values}\mathrm{of}x$$

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