a-2-b-2-c-2-d-2-4-a-b-c-d-R-max-a-3-b-3-c-3-d-3-

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 R

m a x {a^{3} + b^{3} + c^{3} + d^{3}} = ?

Commented by naka3546 last updated on 28/Oct/21

W h i c h i n e q u a l i t y i s u s e d f o r s o l v i n g i t ?

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

m a x = 8

Commented by naka3546 last updated on 27/Oct/21

S h o w y o u r w o r k i n g s, p l e a s e, s i r .

Commented by naka3546 last updated on 27/Oct/21

T h a n k y o u, s i r .

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

i f x_{1} + x_{2} + x_{3} + \dots + x_{n} = s w i t h x_{i} ⩾ 0,

t h e n f o r p > 1

{(x_{1}^{p} + x_{2}^{p} + x_{3}^{p} + \dots + x_{n}^{p})}_{m i n} = n {(\frac{s}{n})}^{p} = \frac{s^{p}}{n^{p - 1}}

{(x_{1}^{p} + x_{2}^{p} + x_{3}^{p} + \dots + x_{n}^{p})}_{m a x} = s^{p}

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

i {d o n}^{'} t k n o w i f i^{'} m u s i n g a k n o w n

i n e q u a l i t y . a c t u a l l y i j u s t u s e m y

b r a i n a n d l o g i c .

{l e t}^{'} s l o o k a t f (x) = x^{p} w i t h p > 1 .

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

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

w e c a n s e e (c e r t a i n l y {i t}^{'} s a l s o e a s y

t o p r o v e) :

f (x_{1} + x_{2}) ⩾ f (x_{1}) + f (x_{2}), i . e .

{(x_{1} + x_{2})}^{p} ⩾ x_{1}^{p} + x_{2}^{p}

(e q u a l i t y h o l d s w h e n x_{2} = 0)

a p p l y i n g t h i s w e g e t

{(x_{1} + x_{2} + x_{3})}^{p} ⩾ {(x_{1} + x_{2})}^{p} + x_{3}^{p} ⩾ x_{1}^{p} + x_{2}^{p} + x_{3}^{p}

a n d \dots

{(x_{1} + x_{2} + x_{3} + \dots + x_{n})}^{p} ⩾ x_{1}^{p} + x_{2}^{p} + x_{3}^{p} + \dots + x_{n}^{p}

\Rightarrow x_{1}^{p} + x_{2}^{p} + x_{3}^{p} + \dots + x_{n}^{p} ⩽ {(x_{1} + x_{2} + x_{3} + \dots + x_{n})}^{p} = s^{p}

i . e .

{(x_{1}^{p} + x_{2}^{p} + x_{3}^{p} + \dots + x_{n}^{p})}_{m a x} = s^{p}

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

b a c k t o t h e q u e s t i o n, w e h a v e

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})}_{m a x} = 8

Commented by naka3546 last updated on 29/Oct/21

T h a n k y o u s o m u c h, s i r .