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

  x , y , z are positive real numbers if x^4 +y^4 +z^4 =1  Then find the minimum value of   (x^3 /(1−x^8 ))+(y^3 /(1−y^8 ))+(z^3 /(1−z^8 ))

$$ \\ $$$${x}\:,\:{y}\:,\:{z}\:{are}\:{positive}\:{real}\:{numbers}\:{if}\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\mathrm{1} \\ $$$${Then}\:{find}\:{the}\:{minimum}\:{value}\:{of}\: \\ $$$$\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{8}} }+\frac{{y}^{\mathrm{3}} }{\mathrm{1}−{y}^{\mathrm{8}} }+\frac{{z}^{\mathrm{3}} }{\mathrm{1}−{z}^{\mathrm{8}} } \\ $$

Answered by Subhi last updated on 27/Jun/23

put f(x) = (x^3 /(1−x^8 ))  f^′ (x) = ((3x^2 (1−x^8 )+8x^7 (x^3 ))/((1−x^8 )^2 )) at x = (1/(^4 (√3))) (value that achieves the equality)  f^′ ((1/(^4 (√3)))) = 2.598  y_f −y_0  = m(x_f −x_0 )  f(x) = 2.598(x−(1/(^4 (√3))))+((((1/(^4 (√3))))^3 )/(1−((1/(^4 (√3))))^8 ))  f(x) = 2.598x −1.48  according to tangent theorem  f(x)+f(y)+f(z) ≥ 2.589(x+y+z)−1.48×3   = 2.589((3/(^4 (√3))))−1.48×3 = 1.48 approx

$${put}\:{f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{8}} } \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\mathrm{3}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{8}} \right)+\mathrm{8}{x}^{\mathrm{7}} \left({x}^{\mathrm{3}} \right)}{\left(\mathrm{1}−{x}^{\mathrm{8}} \right)^{\mathrm{2}} }\:{at}\:{x}\:=\:\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{3}}}\:\left({value}\:{that}\:{achieves}\:{the}\:{equality}\right) \\ $$$${f}^{'} \left(\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{3}}}\right)\:=\:\mathrm{2}.\mathrm{598} \\ $$$${y}_{{f}} −{y}_{\mathrm{0}} \:=\:{m}\left({x}_{{f}} −{x}_{\mathrm{0}} \right) \\ $$$${f}\left({x}\right)\:=\:\mathrm{2}.\mathrm{598}\left({x}−\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{3}}}\right)+\frac{\left(\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{3}}}\right)^{\mathrm{3}} }{\mathrm{1}−\left(\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{3}}}\right)^{\mathrm{8}} } \\ $$$${f}\left({x}\right)\:=\:\mathrm{2}.\mathrm{598}{x}\:−\mathrm{1}.\mathrm{48} \\ $$$${according}\:{to}\:{tangent}\:{theorem} \\ $$$${f}\left({x}\right)+{f}\left({y}\right)+{f}\left({z}\right)\:\geqslant\:\mathrm{2}.\mathrm{589}\left({x}+{y}+{z}\right)−\mathrm{1}.\mathrm{48}×\mathrm{3} \\ $$$$\:=\:\mathrm{2}.\mathrm{589}\left(\frac{\mathrm{3}}{\:^{\mathrm{4}} \sqrt{\mathrm{3}}}\right)−\mathrm{1}.\mathrm{48}×\mathrm{3}\:=\:\mathrm{1}.\mathrm{48}\:{approx}\: \\ $$

Commented by Frix last updated on 28/Jun/23

I think at x=y=z=3^(−(1/4))  the minimum is  3^(9/4) 2^(−3) ≈1.48058326457

$$\mathrm{I}\:\mathrm{think}\:\mathrm{at}\:{x}={y}={z}=\mathrm{3}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is} \\ $$$$\mathrm{3}^{\frac{\mathrm{9}}{\mathrm{4}}} \mathrm{2}^{−\mathrm{3}} \approx\mathrm{1}.\mathrm{48058326457} \\ $$

Commented by York12 last updated on 28/Jun/23

thanks bro

$${thanks}\:{bro} \\ $$

Commented by York12 last updated on 28/Jun/23

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$${sir}\:{I}\:{have}\:{sent}\:{you}\:{a}\:{friend}\:{request}\:{on} \\ $$$${discord} \\ $$

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

sir I have sent you a friend request on   discord    please accept sir

$${sir}\:{I}\:{have}\:{sent}\:{you}\:{a}\:{friend}\:{request}\:{on}\: \\ $$$${discord}\:\: \\ $$$${please}\:{accept}\:{sir} \\ $$

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