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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
sir I have sent you a friend request on discord
$${sir}\:{I}\:{have}\:{sent}\:{you}\:{a}\:{friend}\:{request}\:{on} \\ $$$${discord} \\ $$
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
$${sir}\:{I}\:{have}\:{sent}\:{you}\:{a}\:{friend}\:{request}\:{on}\: \\ $$$${discord}\:\: \\ $$$${please}\:{accept}\:{sir} \\ $$
Answered by AST last updated on 28/Jun/23
≥(1/3)(x^3 +y^3 +z^3 )(Σ(1/(1−x^8 )))≥(1/3)(Σx^3 )((9/(3−Σx^8 ))) (Σ(x^8 /3))^(1/8) ≥(Σ(x^4 /3))^(1/4) =(1/( (3)^(1/4) ))⇒Σx^8 ≥(1/3)⇒−Σx^8 ≤−(1/3) ⇒Σ(x^3 /(1−x^8 ))≥(1/3)(Σx^3 )(((27)/8))=(9/8)(Σx^3 ) It remains to find min(Σx^3 )... Use Lagrange multipliers 3x^2 =λ4x^3 ⇒3=4λx⇒λ=(3/(4x))⇒x=(3/(4λ)) ⇒((3×81)/(256λ^4 ))=1⇒λ=(((3×81)/(256)))^(1/4) =((3(3)^(1/4) )/4)⇒x=y=z=((3/1)/((3(3)^(1/4) )/1))=(1/( (3)^(1/4) )) So,min(x^3 +y^3 +z^3 ) holds when x=y=z=(1/( (3)^(1/4) )) ⇒min(x^3 +y^3 +z^3 )=(3/( ((27))^(1/4) )) ⇒Σ(x^3 /(1−x^8 ))≥(9/8)×(3/( ((27))^(1/4) ))=(((27^3 ))^(1/4) /8)
$$\geqslant\frac{\mathrm{1}}{\mathrm{3}}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)\left(\Sigma\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{8}} }\right)\geqslant\frac{\mathrm{1}}{\mathrm{3}}\left(\Sigma{x}^{\mathrm{3}} \right)\left(\frac{\mathrm{9}}{\mathrm{3}−\Sigma{x}^{\mathrm{8}} }\right) \\ $$$$\left(\Sigma\frac{{x}^{\mathrm{8}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} \geqslant\left(\Sigma\frac{{x}^{\mathrm{4}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} =\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{3}}}\Rightarrow\Sigma{x}^{\mathrm{8}} \geqslant\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow−\Sigma{x}^{\mathrm{8}} \leqslant−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\Sigma\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{8}} }\geqslant\frac{\mathrm{1}}{\mathrm{3}}\left(\Sigma{x}^{\mathrm{3}} \right)\left(\frac{\mathrm{27}}{\mathrm{8}}\right)=\frac{\mathrm{9}}{\mathrm{8}}\left(\Sigma{x}^{\mathrm{3}} \right) \\ $$$${It}\:{remains}\:{to}\:{find}\:{min}\left(\Sigma{x}^{\mathrm{3}} \right)…\:{Use}\:{Lagrange}\:{multipliers} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} =\lambda\mathrm{4}{x}^{\mathrm{3}} \Rightarrow\mathrm{3}=\mathrm{4}\lambda{x}\Rightarrow\lambda=\frac{\mathrm{3}}{\mathrm{4}{x}}\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{4}\lambda} \\ $$$$\Rightarrow\frac{\mathrm{3}×\mathrm{81}}{\mathrm{256}\lambda^{\mathrm{4}} }=\mathrm{1}\Rightarrow\lambda=\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}×\mathrm{81}}{\mathrm{256}}}=\frac{\mathrm{3}\sqrt[{\mathrm{4}}]{\mathrm{3}}}{\mathrm{4}}\Rightarrow{x}={y}={z}=\frac{\frac{\mathrm{3}}{\mathrm{1}}}{\frac{\mathrm{3}\sqrt[{\mathrm{4}}]{\mathrm{3}}}{\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{3}}} \\ $$$${So},{min}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)\:{holds}\:{when}\:{x}={y}={z}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{3}}} \\ $$$$\Rightarrow{min}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)=\frac{\mathrm{3}}{\:\sqrt[{\mathrm{4}}]{\mathrm{27}}} \\ $$$$\Rightarrow\Sigma\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{8}} }\geqslant\frac{\mathrm{9}}{\mathrm{8}}×\frac{\mathrm{3}}{\:\sqrt[{\mathrm{4}}]{\mathrm{27}}}=\frac{\sqrt[{\mathrm{4}}]{\mathrm{27}^{\mathrm{3}} }}{\mathrm{8}} \\ $$
Commented by York12 last updated on 28/Jun/23
thanks sir
$${thanks}\:{sir} \\ $$

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