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Question Number 85854 by jagoll last updated on 25/Mar/20

If x,y,z ∈ R satisfy the equation  x^4  + y^4  + z^4  = 4xyz −1   find minimum value of  x + y + z

$$\mathrm{If}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{R}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{y}^{\mathrm{4}} \:+\:\mathrm{z}^{\mathrm{4}} \:=\:\mathrm{4xyz}\:−\mathrm{1}\: \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\: \\ $$

Commented by mr W last updated on 25/Mar/20

seems to be 3

$${seems}\:{to}\:{be}\:\mathrm{3} \\ $$

Commented by mr W last updated on 25/Mar/20

x=y=z=1 fulfills the symmetry and  x^4 +y^4 +z^4 =4xyz−1  ⇒(x+y+z)_(min) =1+1+1=3

$${x}={y}={z}=\mathrm{1}\:{fulfills}\:{the}\:{symmetry}\:{and} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\mathrm{4}{xyz}−\mathrm{1} \\ $$$$\Rightarrow\left({x}+{y}+{z}\underset{{min}} {\right)}=\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$

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