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Question Number 181438 by Agnibhoo98 last updated on 26/Nov/22

$$\mathrm{If}a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}\mathrm{then}\mathrm{prove}\mathrm{that}$$$$abc=\pm 1.a\ne b\ne c$$

Commented by mr W last updated on 26/Nov/22

$${it}^{\prime }snottrue.e.g.youcantake$$$$a=b=c=2,andgetabc=8.$$

Answered by manxsol last updated on 26/Nov/22

$$havef(a,b,c)$$$$restriccions$$$$a+\frac{1}{b}=k\Rightarrow g\left(\right)=a+\frac{1}{b}-k=0$$$$b+\frac{1}{c}=k\Rightarrow h\left(\right)=b+\frac{1}{c}-k=0$$$$c+\frac{1}{a}=k\Rightarrow j\left(\right)=c+\frac{1}{\mathrm{a}}-\mathrm{k}=0$$$$▽g\left(\right)=(\frac{\partial g}{\partial x};\frac{\partial g}{\partial y};\frac{\partial g}{\partial z})=(g_x;g_y;g_z)$$$$df={\lambda }_1▽g+{\lambda }_2▽h+{\lambda }_3▽j=0$$$$df(a,b,c,{\lambda }_1,{\lambda }_2,{\lambda }_3)=$$$${\lambda }_1(1,-\frac{1}{b^2},0)+{\lambda }_2(0,1,-\frac{1}{c^2})+{\lambda }_3(-\frac{1}{a^2},0,1)=(0,0,0)$$$$\left(\begin{array}{cccc}{\lambda }_1& {\lambda }_2& {\lambda }_3& const\\ 1& 0& -\frac{1}{a^2}& 0\\ -\frac{1}{b^2}& 1& 0& 0\\ 0& -\frac{1}{c^2}& 1& 0\end{array}\right)$$$$\frac{{\lambda }_2}{c^2}={\lambda }_3\Rightarrow {\lambda }_2=c^2{\lambda }_3$$$$\frac{{\lambda }_1}{b^2}={\lambda }_2\Rightarrow {\lambda }_1=b^2{\lambda }_2$$$${\lambda }_1=\frac{{\lambda }_3}{a^2}\Rightarrow {\lambda }_3=a^2{\lambda }_1$$$$multiplicationexprexions$$$${\lambda }_1{\lambda }_2{\lambda }_3=a^2{b}^2{c}^2{\lambda }_1{\lambda }_2{\lambda }_3$$$$1=a^2{b}^2{c}^2$$$$abc=\pm 1$$$$LQQD$$$$$$$$$$

Commented by mr W last updated on 26/Nov/22

$$whatisproved?$$$$with2conditionsfor3variables$$$$abccangetanyvalue.$$

Commented by manxsol last updated on 26/Nov/22

$$excuseme,Sr.W.howshouldIdoit$$

Commented by manxsol last updated on 26/Nov/22

$$ortheconditionismissinga\ne b\ne c$$

Commented by Agnibhoo98 last updated on 26/Nov/22

$$\mathrm{yes}$$

Commented by manxsol last updated on 26/Nov/22

$$Sr.W,myprocedureiscorrect???$$

Commented by mr W last updated on 26/Nov/22

$$somethingiswronginthequestion.$$$$anadditionalconditionisneeded.$$$$a\ne b\ne ccouldbeasuchcondition.$$

Commented by mr W last updated on 26/Nov/22

$$i{can}^{\prime }tjudge,becausei{don}^{\prime }tknow$$$$whatyouwanttoprove.$$

Commented by manxsol last updated on 26/Nov/22

$$iunderstand,Iwillreflecton[thequestion.thankyouverymuchforyouattention$$

Answered by Agnibhoo98 last updated on 27/Nov/22

$$a+\frac{1}{b}=b+\frac{1}{c}$$$$ora-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}....\left(1\right)$$$$b+\frac{1}{c}=c+\frac{1}{a}$$$$orb-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ca}....\left(2\right)$$$$a+\frac{1}{b}=c+\frac{1}{a}$$$$ora-c=\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}....\left(3\right)$$$$\mathrm{Multiplying}\left(1\right),\left(2\right),\left(3\right)$$$$$$$$(a-b)(b-c)(a-c)=\frac{(b-c)(c-a)(b-a)}{{\left(abc\right)}^2}$$$$or{\left(abc\right)}^2=\frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(a-c)}$$$$or{\left(abc\right)}^2=1$$$$orabc=\pm 1$$$$$$

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