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Question-207730




Question Number 207730 by mr W last updated on 24/May/24
Commented by mr W last updated on 24/May/24
is this possible?
$${is}\:{this}\:{possible}? \\ $$
Commented by efronzo1 last updated on 25/May/24
i think , no sir
$$\mathrm{i}\:\mathrm{think}\:,\:\mathrm{no}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 25/May/24
i believe it is possible. reason:  x_1 ^2 +x_2 ^2 +x_3 ^2 +...+x_(2023) ^2 =125  is a sphere in a 2023−dimensional  space with radius (√(125)).  x_1 +x_2 +x_3 +...+x_(2023) =25  is a plane in this 2023−dimensional  space.  we can see that the plane and  the sphere intersect each other.   the intersection is a ring in that   space. at least one point on this ring  delivers the maximum value for  x_1 ^3 +x_2 ^3 +x_3 ^3 +...+x_(2023) ^3  and at least one   point delivers the minimum value.
$${i}\:{believe}\:{it}\:{is}\:{possible}.\:{reason}: \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +{x}_{\mathrm{3}} ^{\mathrm{2}} +…+{x}_{\mathrm{2023}} ^{\mathrm{2}} =\mathrm{125} \\ $$$${is}\:{a}\:{sphere}\:{in}\:{a}\:\mathrm{2023}−{dimensional} \\ $$$${space}\:{with}\:{radius}\:\sqrt{\mathrm{125}}. \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +…+{x}_{\mathrm{2023}} =\mathrm{25} \\ $$$${is}\:{a}\:{plane}\:{in}\:{this}\:\mathrm{2023}−{dimensional} \\ $$$${space}. \\ $$$${we}\:{can}\:{see}\:{that}\:{the}\:{plane}\:{and} \\ $$$${the}\:{sphere}\:{intersect}\:{each}\:{other}.\: \\ $$$${the}\:{intersection}\:{is}\:{a}\:{ring}\:{in}\:{that}\: \\ $$$${space}.\:{at}\:{least}\:{one}\:{point}\:{on}\:{this}\:{ring} \\ $$$${delivers}\:{the}\:{maximum}\:{value}\:{for} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{3}} +{x}_{\mathrm{2}} ^{\mathrm{3}} +{x}_{\mathrm{3}} ^{\mathrm{3}} +…+{x}_{\mathrm{2023}} ^{\mathrm{3}} \:{and}\:{at}\:{least}\:{one}\: \\ $$$${point}\:{delivers}\:{the}\:{minimum}\:{value}. \\ $$

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