Question-195454

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 AST 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

Commented by AST last updated on 03/Aug/23

I t s e e m s s o .

Answered by witcher3 last updated on 03/Aug/23

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 \dots \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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