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Question Number 195454 by universe last updated on 02/Aug/23

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

$s e e a l s o Q 157749$
	Answered by Frix last updated on 03/Aug/23

	$a = 2 b = c = 0$ $2^{3} = 8$
	Commented by universe last updated on 03/Aug/23

	$i k n o w t h i s m e t h o d$ $b u t i w a n t s o l u t i o n i n m a t h e m a t i c s f o r m$
	Answered by witcher3 last updated on 03/Aug/23

	$use \sqrt[q]{\frac{\underset{i = 1}{\sum^{n}} x_{i}^{q}}{n}} = f (q) increse function$ $\Rightarrow \sqrt[3]{\frac{a^{3} + b^{3} + c^{3}}{3}} ⩾ {(\frac{a^{2} + b^{2} + c^{2}}{3})}^{\frac{1}{2}} = \frac{2}{\sqrt{3}}$ $\Rightarrow a^{3} + b^{3} + c^{2} ⩾ \frac{8}{3 \sqrt{3}} . 3 = \frac{8}{\sqrt{3}}$ $a = b = c \Rightarrow \frac{2}{\sqrt{3}}$
	Commented by deleteduser1 last updated on 03/Aug/23

	$T h i s i s m i n i m u m v a l u e .$
	Commented by witcher3 last updated on 03/Aug/23

	$yes i dont know why in my mind this is the Quation$
	Answered by witcher3 last updated on 03/Aug/23
	$a^3 +b^3 +c^3 =4((a^3 /(a^2 +b^2 +c^2 ))+(b^3 /(a^2 +b^2 +c^2 ))+(c^3 /(a^2 +b^2 +c^3 ))) =4(((a^3 +b^3 +c^3 )/(b^2 +c^2 +a^2 ))) M∈R_+ ,a^3 +b^3 +c^3 ≤M(a^2 +b^2 +c^2 ) M=max {a,b,c}=a;a^3 +b^3 +c^3 ≤a^3 +a(b^2 +c^2 ) true if a≥1 a^2 +b^2 +c^2 ≤3max{a^2 ,b^2 ,c^2 } max{a^2 ,b^2 ,c^2 }=max^2 {a,b,c}≥(4/3)⇒m≥(2/( (√3)))>1.. True ,hence Get idea if one of a,b,c Get bigger a^3 +b^3 +c^3 get bigger a=2,b=c=0 a^3 +b^3 +c^3 =8 a^3 +b^3 +c^3 ≤8...∀(a,b,c)∣a^2 +b^2 +c^2 =4⇒(a,b,c)∈[0,2]^2 8=2(a^2 +b^2 +c^2 ) a^3 +b^3 +c^3 =a.a^2 +b.b^2 +c.c^2 ≤2a^2 +2b^2 +2c^2 =2(a^2 +b^2 +c^2 )=8$
	$a^{3} + b^{3} + c^{3} = 4 (\frac{a^{3}}{a^{2} + b^{2} + c^{2}} + \frac{b^{3}}{a^{2} + b^{2} + c^{2}} + \frac{c^{3}}{a^{2} + b^{2} + c^{3}})$ $= 4 (\frac{a^{3} + b^{3} + c^{3}}{b^{2} + c^{2} + a^{2}})$ $M \in R_{+}, a^{3} + b^{3} + c^{3} ⩽ M (a^{2} + b^{2} + c^{2})$ $M = \max {a, b, c} = a; a^{3} + b^{3} + c^{3} ⩽ a^{3} + a (b^{2} + c^{2})$ $true if a ⩾ 1$ $a^{2} + b^{2} + c^{2} ⩽ 3 \max {a^{2}, b^{2}, c^{2}}$ $\max {a^{2}, b^{2}, c^{2}} = \max^{2} {a, b, c} ⩾ \frac{4}{3} \Rightarrow m ⩾ \frac{2}{\sqrt{3}} > 1 . .$ $True, hence Get idea if one of a, b, c Get bigger$ $a^{3} + b^{3} + c^{3} get bigger$ $a = 2, b = c = 0$ $a^{3} + b^{3} + c^{3} = 8$ $a^{3} + b^{3} + c^{3} ⩽ 8 . . . \forall (a, b, c) ∣ a^{2} + b^{2} + c^{2} = 4 \Rightarrow (a, b, c) \in {[0, 2]}^{2}$ $8 = 2 (a^{2} + b^{2} + c^{2})$ $a^{3} + b^{3} + c^{3} = a . a^{2} + b . b^{2} + c . c^{2} ⩽ {2 a}^{2} + {2 b}^{2} + {2 c}^{2} = 2 (a^{2} + b^{2} + c^{2}) = 8$
	Answered by universe last updated on 04/Aug/23

	$L E T$ $a = 2 \sin α \cos β$ $b = 2 \sin α \sin β$ $c = 2 \cos α$ $a^{3} + b^{3} + c^{3} = 8 \underset{m a x = 1}{\underset{⏟}{[\sin^{3} α (\cos^{3} β + \sin^{3} β) + \cos^{3} α]}}$ $m a x (a^{3} + b^{3} + c^{3}) = 8$
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