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Question Number 157749 by naka3546 last updated on 27/Oct/21

$$a^2+b^2+c^2+d^2=4$$$$a,b,c,d\in \mathbb{R}$$$$max\{a^3+b^3+c^3+d^3\}=?$$

Commented by naka3546 last updated on 28/Oct/21

$$\mathrm{W}hichinequalityisusedforsolvingit?$$

Commented by mr W last updated on 27/Oct/21

$$max=8$$

Commented by naka3546 last updated on 27/Oct/21

$$Showyourworkings,please,sir.$$

Commented by naka3546 last updated on 27/Oct/21

$$Thankyou,sir.$$

Commented by mr W last updated on 27/Oct/21

$$if{x}_1+x_2+x_3+...+x_n=swith{x}_i⩾0,$$$$thenforp>1$$$${(x_1^p+x_2^p+x_3^p+...+x_n^p)}_{min}=n{\left(\frac{s}{n}\right)}^p=\frac{s^p}{n^{p-1}}$$$${(x_1^p+x_2^p+x_3^p+...+x_n^p)}_{max}=s^p$$

Commented by mr W last updated on 03/Aug/23

$$i{don}^{\prime }tknowif{i}^{\prime }musingaknown$$$$inequality.actuallyijustusemy$$$$brainandlogic.$$$${let}^{\prime }slookatf\left(x\right)=x^{p}withp>1.$$

Commented by mr W last updated on 28/Oct/21

Commented by mr W last updated on 03/Aug/23

$$wecansee(certainly{it}^{\prime }salsoeasy$$$$toprove):$$$$f(x_1+x_2)⩾f\left(x_1\right)+f\left(x_2\right),i.e.$$$${(x_1+x_2)}^p⩾x_1^p+x_2^p$$$$(equalityholdswhen{x}_2=0)$$$$applyingthisweget$$$${(x_1+x_2+x_3)}^p⩾{(x_1+x_2)}^p+x_3^p⩾x_1^p+x_2^p+x_3^p$$$$and...$$$${(x_1+x_2+x_3+...+x_n)}^p⩾x_1^p+x_2^p+x_3^p+...+x_n^p$$$$\Rightarrow x_1^p+x_2^p+x_3^p+...+x_n^p⩽{(x_1+x_2+x_3+...+x_n)}^p=s^p$$$$i.e.$$$${(x_1^p+x_2^p+x_3^p+...+x_n^p)}_{max}=s^p$$

Commented by mr W last updated on 28/Oct/21

$$backtothequestion,wehave$$$$x_1=a^2,x_2=b^2,x_3=c^2,x_4=d^2$$$$s=a^2+b^2+c^2+d^2=x_1+x_2+x_3+x_4=4$$$$p=\frac{3}{2}>1$$$$a^3+b^3+c^3+d^3=x_1^p+x_2^p+x_3^p+x_4^p⩽4^p=4^{\frac{3}{2}}=8$$$$\Rightarrow {(a^3+b^3+c^3+d^3)}_{max}=8$$

Commented by naka3546 last updated on 29/Oct/21

$$Thankyousomuch,sir.$$

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