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


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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

$oo yes it my typo$
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