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Question Number 107794 by udaythool last updated on 12/Aug/20

Answered by arcana last updated on 12/Aug/20

$$\mathrm{Considere}$$$$f\left(x\right)=x\sqrt{1-x^2}$$$$\Rightarrow f{}^{\prime }\left(x\right)=\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=0,$$$$\mathrm{luego}1-2x^2=0\Rightarrow x=\frac{1}{\sqrt{2}}.$$$$\mathrm{Decimos}\mathrm{que}(1/\sqrt{2},f\left(1/\sqrt{2}\right))\mathrm{es}\mathrm{punto}$$$$\mathrm{maximo}\mathrm{de}f.$$$$\mathrm{Por}\mathrm{tanto},f\left(x\right)⩽f\left(1/\sqrt{2}\right).$$$$\mathrm{Como}\mid z\mid =x^2+y^2⩽1,\mathrm{entonces}$$$$xy⩽x\sqrt{1-x^2}⩽1/2.$$$$$$

Commented by udaythool last updated on 13/Aug/20

$$\mathrm{Excellent}\mathrm{attempt},\mathrm{but}\mathrm{it}\mathrm{does}$$$$\mathrm{not}\mathrm{guarentee}\mathrm{that}\mid xy\mid ⩽\frac{1}{2}$$

Commented by udaythool last updated on 14/Aug/20

$$\mathrm{I}\mathrm{think}\mathrm{by}\mathrm{contradiction}\mathrm{method}$$$$\mathrm{we}\mathrm{can}\mathrm{prove}\mathrm{it}\mathrm{easily};$$$$\mathrm{Suppose}\mathrm{\forall }z=x+iy\in A,\mid xy\mid ⪇\frac{1}{2}$$$$\Rightarrow \mathrm{\exists }\mathrm{some}z=x+iy\in A\mathrm{such}\mathrm{that}$$$$\frac{1}{2}<\mid xy\mid$$$$\mathrm{But}{x}^2+y^2⩽1\mathrm{hence}$$$$0⩾x^2+y^2-1>{(\frac{1}{2y}-y)}^2⩾0$$$$\mathrm{which}\mathrm{is}\mathrm{absurd}...$$

Answered by mr W last updated on 12/Aug/20

$$\mid z\mid =\sqrt{x^2+y^2}⩽1$$$$\Rightarrow x^2+y^2⩽1$$$$\Rightarrow x^2+y^2-2xy+2xy⩽1$$$$\Rightarrow {(x-y)}^2+2xy⩽1$$$$\Rightarrow xy⩽\frac{1}{2}-\frac{{(x-y)}^2}{2}⩽\frac{1}{2}$$

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