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Question Number 194105 by York12 last updated on 27/Jun/23

$$$$$$x,y,zarepositiverealnumbersif{x}^4+y^4+z^4=1$$$$Thenfindtheminimumvalueof$$$$\frac{x^3}{1-x^8}+\frac{y^3}{1-y^8}+\frac{z^3}{1-z^8}$$

Answered by Subhi last updated on 27/Jun/23

$$putf\left(x\right)=\frac{x^3}{1-x^8}$$$$f^{{}^{\prime }}\left(x\right)=\frac{3x^2(1-x^8)+8x^7\left(x^3\right)}{{(1-x^8)}^2}atx=\frac{1}{{}^4\sqrt{3}}\left(valuethatachievestheequality\right)$$$$f^{{}^{\prime }}\left(\frac{1}{{}^4\sqrt{3}}\right)=2.598$$$$y_f-y_0=m(x_f-x_0)$$$$f\left(x\right)=2.598(x-\frac{1}{{}^4\sqrt{3}})+\frac{{\left(\frac{1}{{}^4\sqrt{3}}\right)}^3}{1-{\left(\frac{1}{{}^4\sqrt{3}}\right)}^8}$$$$f\left(x\right)=2.598x-1.48$$$$accordingtotangenttheorem$$$$f\left(x\right)+f\left(y\right)+f\left(z\right)⩾2.589(x+y+z)-1.48\times 3$$$$=2.589\left(\frac{3}{{}^4\sqrt{3}}\right)-1.48\times 3=1.48approx$$

Commented by Frix last updated on 28/Jun/23

$$\mathrm{I}\mathrm{think}\mathrm{at}x=y=z=3^{-\frac{1}{4}}\mathrm{the}\mathrm{minimum}\mathrm{is}$$$$3^{\frac{9}{4}}{2}^{-3}\approx 1.48058326457$$

Commented by York12 last updated on 28/Jun/23

$$thanksbro$$

Commented by York12 last updated on 28/Jun/23

$$sirIhavesentyouafriendrequeston$$$$discord$$

Commented by York12 last updated on 28/Jun/23

Commented by York12 last updated on 28/Jun/23

Commented by York12 last updated on 28/Jun/23

$$sirIhavesentyouafriendrequeston$$$$discord$$$$pleaseacceptsir$$

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