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Question Number 195570 by York12 last updated on 05/Aug/23

Given three Real numbers (x,y,z),such that  x^2 +y^2 +z^2 =1  maximize  x^4 +y^4 −2z^4 −3(√2)xyz

$$\mathrm{Given}\:\mathrm{three}\:\mathrm{Real}\:\mathrm{numbers}\:\left({x},{y},{z}\right),{such}\:{that} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1} \\ $$$${maximize} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{4}} −\mathrm{3}\sqrt{\mathrm{2}}{xyz} \\ $$

Commented by Frix last updated on 05/Aug/23

I think the maximum is ((65)/(64)) when z=±((√2)/4)

$$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{is}\:\frac{\mathrm{65}}{\mathrm{64}}\:\mathrm{when}\:{z}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Commented by York12 last updated on 05/Aug/23

please provide workings

$${please}\:{provide}\:{workings} \\ $$

Answered by Frix last updated on 05/Aug/23

x^2 +y^2 =1−z^2  ⇒ x^4 +y^4 =−2x^2 y^2 +z^4 −2z^2 +1    x^4 +y^4 −2z^4 −3(√2)xyz=  =−2x^2 y^2 −3(√2)xyz−z^4 −2z^2 +1  Let t=xy  −2t^2 −2(√2)tz−z^4 −2z^2 +1  ((d[−2t^2 −2(√2)tz−z^4 −2z^2 +1])/dt)=0  −4t−3(√2)z=0 ⇒ t=−((3(√2)z)/4)  Inserting  −z^4 +(z^2 /4)+1  ((d[−z^4 +(z^2 /4)+1])/dz)=0  −4z^3 +(z/2)=0 ⇒ z=±((√2)/4) [∨z=0_(minimum) ^(rejected because) ]  ⇒ maximum is ((65)/(64)) ★  We have  z=((√2)/4)∧xy=−(3/8) ∨ z=−((√2)/4)∧xy=(3/8)  x^2 +y^2 =1−z^2 =(7/8)  The values are easy to calculate

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}−{z}^{\mathrm{2}} \:\Rightarrow\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =−\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{1} \\ $$$$ \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{4}} −\mathrm{3}\sqrt{\mathrm{2}}{xyz}= \\ $$$$=−\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}}{xyz}−{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{Let}\:{t}={xy} \\ $$$$−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{tz}−{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{1} \\ $$$$\frac{{d}\left[−\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}{tz}−{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{2}} +\mathrm{1}\right]}{{dt}}=\mathrm{0} \\ $$$$−\mathrm{4}{t}−\mathrm{3}\sqrt{\mathrm{2}}{z}=\mathrm{0}\:\Rightarrow\:{t}=−\frac{\mathrm{3}\sqrt{\mathrm{2}}{z}}{\mathrm{4}} \\ $$$$\mathrm{Inserting} \\ $$$$−{z}^{\mathrm{4}} +\frac{{z}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1} \\ $$$$\frac{{d}\left[−{z}^{\mathrm{4}} +\frac{{z}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}\right]}{{dz}}=\mathrm{0} \\ $$$$−\mathrm{4}{z}^{\mathrm{3}} +\frac{{z}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\:{z}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\left[\vee{z}=\mathrm{0}_{\mathrm{minimum}} ^{\mathrm{rejected}\:\mathrm{because}} \right] \\ $$$$\Rightarrow\:\mathrm{maximum}\:\mathrm{is}\:\frac{\mathrm{65}}{\mathrm{64}}\:\bigstar \\ $$$$\mathrm{We}\:\mathrm{have} \\ $$$$\mathrm{z}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\wedge{xy}=−\frac{\mathrm{3}}{\mathrm{8}}\:\vee\:{z}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\wedge{xy}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}−{z}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\mathrm{The}\:\mathrm{values}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{calculate} \\ $$

Commented by York12 last updated on 05/Aug/23

ORZ

$${ORZ} \\ $$$$ \\ $$$$ \\ $$

Commented by York12 last updated on 05/Aug/23

where to learn inequalities   and are you using discord   I would like to invite you to our community

$${where}\:{to}\:{learn}\:{inequalities}\: \\ $$$${and}\:{are}\:{you}\:{using}\:{discord}\: \\ $$$${I}\:{would}\:{like}\:{to}\:{invite}\:{you}\:{to}\:{our}\:{community} \\ $$$$ \\ $$

Commented by Frix last updated on 05/Aug/23

Sorry I don't use any social media platforms. I studied mathematics & physics decades ago at a university, I don't know where else you could learn these things.

Commented by York12 last updated on 05/Aug/23

okay sir thanks

$${okay}\:{sir}\:{thanks} \\ $$

Commented by York12 last updated on 05/Aug/23

but if you wanted you are always welcome  we are a big community of high scholars  from different spots on the earth and we  need instructions from an efficient  person like you sir

$${but}\:{if}\:{you}\:{wanted}\:{you}\:{are}\:{always}\:{welcome} \\ $$$${we}\:{are}\:{a}\:{big}\:{community}\:{of}\:{high}\:{scholars} \\ $$$${from}\:{different}\:{spots}\:{on}\:{the}\:{earth}\:{and}\:{we} \\ $$$${need}\:{instructions}\:{from}\:{an}\:{efficient} \\ $$$${person}\:{like}\:{you}\:{sir} \\ $$

Commented by necx122 last updated on 06/Aug/23

you could share the link, a few of us are  also interested.

$${you}\:{could}\:{share}\:{the}\:{link},\:{a}\:{few}\:{of}\:{us}\:{are} \\ $$$${also}\:{interested}. \\ $$

Commented by York12 last updated on 06/Aug/23

great

$${great} \\ $$

Commented by York12 last updated on 06/Aug/23

  https://discord.gg/mods

$$ \\ $$https://discord.gg/mods

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