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Question Number 82375 by M±th+et£s last updated on 20/Feb/20

$$ifx+y=8,,x,y\in {\mathbb{R}}^{+}$$$$provethat$$$${(x+\frac{1}{y})}^2+{(y+\frac{1}{x})}^2⩾\frac{289}{8}$$

Answered by MJS last updated on 21/Feb/20

$$f\left(x\right)={(x+\frac{1}{8-x})}^2+{(8-x+\frac{1}{x})}^2$$$$f\left(x\right)=\frac{2(x^2-8x-1)(x^2-8x+32)}{x^2{(x-8)}^2}$$$$\mathrm{shift}4\mathrm{to}\mathrm{the}\mathrm{left}$$$$x=t+4\iff t=x-4$$$$f\left(t\right)=\frac{2{(t^2-17)}^2(t^2+16)}{{(t-4)}^2{(t+4)}^2}$$$$f\left(t\right)\mathrm{has}\mathrm{double}\mathrm{zeros}\mathrm{at}t=\pm \sqrt{17}\mathrm{which}\mathrm{also}$$$$\mathrm{are}\mathrm{minima}$$$$\mathrm{and}\mathrm{a}\mathrm{local}\mathrm{minimum}\mathrm{at}t=0\mathrm{with}f\left(0\right)=\frac{289}{8}$$$$\Rightarrow f\left(x\right)\mathrm{has}\mathrm{double}\mathrm{zeros}\mathrm{at}x=4\pm \sqrt{17}\mathrm{which}$$$$\mathrm{also}\mathrm{are}\mathrm{minima}\mathrm{but}0<x<8\mathrm{so}\mathrm{they}$$$$\mathrm{are}\mathrm{outside}\mathrm{the}\mathrm{given}\mathrm{interval}$$$$\mathrm{and}\mathrm{a}\mathrm{local}\mathrm{minimum}\mathrm{at}x=4\mathrm{with}f\left(4\right)=\frac{289}{8}$$$${\mathrm{what}}^{\prime }\mathrm{s}\mathrm{left}\mathrm{to}\mathrm{show}\mathrm{is}\mathrm{that}f\left(x\right)⩾\frac{289}{8}\mathrm{at}\mathrm{the}$$$$\mathrm{borders}:\mathrm{but}\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 8}{\lim }f\left(x\right)=+\mathrm{\infty }\Rightarrow$$$$\Rightarrow \mathrm{proven}$$

Answered by mind is power last updated on 21/Feb/20

$$x^2+y^2+\frac{1}{y^2}+\frac{1}{x^2}+2\frac{x}{y}+2\frac{y}{x}$$$$=(x^2+y^2)(1+\frac{1}{{\mathrm{x}}^2{\mathrm{y}}^2}+\frac{2}{{\mathrm{xy}}^{}})$$$$,x^2+y^2⩾\frac{{(x+y)}^2}{2}=\frac{64}{2}=\frac{256}{8}$$$$AMGM\Rightarrow$$$$xy⩽\frac{{(x+y)}^2}{4}=16$$$$\Rightarrow \frac{1}{xy}⩾\frac{1}{16}\Rightarrow$$$$(1+\frac{2}{xy}+\frac{1}{x^2{y}^2})⩾1+\frac{2}{16}+\frac{1}{256}=\frac{1+32+256}{256}=\frac{289}{256}$$$$(x^2+y^2)(1+\frac{2}{xy}+\frac{1}{x^2{y}^2})⩾\frac{289}{256}.\frac{256}{8}=\frac{289}{8}$$$$\iff {(x+\frac{1}{y})}^2+{(y+\frac{1}{x})}^2⩾\frac{289}{8}$$$$$$

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