given-a-b-c-a-c-b-b-c-a-9-a-2-b-2-c-2-12-maximum-value-of-a-b-c-is-

Question Number 81564 by jagoll last updated on 14/Feb/20

g i v e n

\frac{a + b}{c} + \frac{a + c}{b} + \frac{b + c}{a} = 9

a^{2} + b^{2} + c^{2} = 12

m a x i m u m v a l u e o f a + b + c i s

Commented by mr W last updated on 14/Feb/20

a + b + c ⩽ 6 \sqrt{3} ?

Commented by john santu last updated on 14/Feb/20

let a + b + c = k

\frac{a + b + c - c}{c} + \frac{a + b + c - b}{b} + \frac{a + b + c - a}{a} = 9

\frac{k}{c} + \frac{k}{b} + \frac{k}{a} = 12 \Rightarrow

k (\frac{b + c}{bc} + \frac{1}{a}) = 12

k (\frac{ab + bc + ac}{abc}) = 12

let ab + bc + ac = m

mk = 12 abc

(i) a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (ab + ac + bc)

ab + ac + bc = \frac{k^{2} - 12}{2}

(ii) k (\frac{k^{2} - 12}{2 abc}) = 12 \Rightarrow k^{3} - 12 k = 24 abc ⩽ 24 {(\frac{k}{3})}^{3}

{27 k}^{3} - 324 k ⩽ {24 k}^{3}

3 k (k^{2} - 108) ⩽ 0

3 k (k + 6 \sqrt{3}) (k - 6 \sqrt{3}) ⩽ 0

for k > 0 \Rightarrow 0 < k ⩽ 6 \sqrt{3}

Commented by mr W last updated on 14/Feb/20

y o u n e e d o n l y a l i t t l e s t e p t o t h e f i n a l

s o l u t i o n :

k^{3} - 12 k = 24 abc ⩽ 24 {(\frac{a + b + c}{3})}^{3} = \frac{24 k^{3}}{27}

\Rightarrow \frac{k^{3}}{9} - 12 k ⩽ 0

\Rightarrow k ⩽ \sqrt{9 \times 12} = 6 \sqrt{3}

Commented by mr W last updated on 14/Feb/20

p l e a s e d i s c u s s :

c a n a + b + c r e a c h t h i s m a x i m u m 6 \sqrt{3} ?

i m e a n, i s a + b + c < 6 \sqrt{3} o r

a + b + c ⩽ 6 \sqrt{3} c o r r e c t ?

Commented by john santu last updated on 14/Feb/20

we use AM sir ? yes i agree with you

Commented by mr W last updated on 14/Feb/20

m y q u e s t i o n i s i f t h e “ =^{″} s i g n i n

a + b + c ⩽ 6 \sqrt{3} i s v a l i d . i t h i n k i t {i s n}^{'} t .

i . e . c o r r e c t i s :

a + b + c < 6 \sqrt{3} .

{w h a t}^{'} s y o u r o p i n i o n ?

Commented by jagoll last updated on 14/Feb/20

i n o t f i n d t h i s s o l u t i o n s i r .

Commented by mr W last updated on 14/Feb/20

y o u m e a n t h i s s o l u t i o n i s w r o n g o r

y o u m e a n y o u {d o n}^{'} t k n o w t h e c o r r e c t

s o l u t i o n ?

Commented by jagoll last updated on 14/Feb/20

i t h i n k t h i s s o l u t i o n i s c o r r e c t .

b u t i {d o n}^{'} t k n o w t h e s o l u t i o n d o e s

i t m a t c h i n t h e b o o k ? a n s w e r

i n t h e b o o k w e r e n o t i n c l u d e d

Commented by mr W last updated on 14/Feb/20

k^{3} - 12 k = 24 abc ⩽ {(\frac{k}{3})}^{3} i s w r o n g .

c o r r e c t i s :

k^{3} - 12 k = 24 abc ⩽ 24 {(\frac{k}{3})}^{3} .

t h i s i s f o r k > 0 .

i f k < 0, t h e n

k^{3} - 12 k = 24 abc ⩾ - 24 {(\frac{k}{3})}^{3} .

Commented by john santu last updated on 14/Feb/20

oo yes it my typo