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Question Number 59409 by hovea cw last updated on 09/May/19

Commented by hovea cw last updated on 09/May/19

$$\mathrm{hplz}\mathrm{help}\mathrm{meelp}\mathrm{plz}$$

Commented by MJS last updated on 10/May/19

$$x=1y=4z=9$$$${3\mathrm{e}}^{14}⩾{\mathrm{e}}^6+{\mathrm{e}}^9+{\mathrm{e}}^{18}$$$$3⩾{\mathrm{e}}^{-8}+{\mathrm{e}}^{-5}+{\mathrm{e}}^4$$$$\mathrm{false}$$$$\mathrm{let}x⩽y\wedge x⩽z$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{false}\mathrm{for}$$$${\mathrm{e}}^{3\sqrt{yz}-(x+y+z)}⩾3$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{false}\mathrm{whenever}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{an}x\mathrm{with}$$$$0<x⩽3\sqrt{yz}-y-z-\ln 3$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{false}\mathrm{when}$$$$0<3\sqrt{yz}-y-z-\ln 3$$$$3\sqrt{yz}-y-z-\ln 3=0$$$$z=\frac{1}{4}{(\sqrt{5y-4\ln 3}\pm 3\sqrt{y})}^2\Rightarrow y⩾\frac{4\ln 3}{5}$$$$\Rightarrow {\mathrm{it}}^{\prime }\mathrm{s}\mathrm{false}\mathrm{when}$$$$y>\frac{4\ln 3}{5}\wedge \frac{1}{4}{(\sqrt{5y-4\ln 3}-3\sqrt{y})}^2<z<\frac{1}{4}{(\sqrt{5y-4\ln 3}+3\sqrt{y})}^2$$

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