a,b,c>0,if ((8a^2 )/(a^2 +9))=b, ((10b^2 )/(b^2 +16))=c, ((6c^2 )/(c^2 +25))=a, then,what is the minimum value of a+b+c?


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Question Number 39868 by math2018 last updated on 12/Jul/18

$a, b, c > 0, i f$ $\frac{8 a^{2}}{a^{2} + 9} = b, \frac{10 b^{2}}{b^{2} + 16} = c, \frac{6 c^{2}}{c^{2} + 25} = a,$ $t h e n, w h a t i s t h e m i n i m u m v a l u e o f a + b + c ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

	$b = \frac{8}{1 + \frac{9}{a^{2}}} = \frac{8}{{(1 - \frac{3}{a})}^{2} + \frac{6}{a}}$ $i f a = 3 t h e n b = 4$ $c = \frac{10}{{(1 - \frac{4}{b})}^{2} + \frac{8}{b}} i f b = 4 c = 5$ $a = \frac{6}{{(1 - \frac{5}{c})}^{2} + \frac{10}{c}} i f c = 5 a = 3$ $s o a = 3$ $b = 4$ $c = 5 s a t i s f y t h r e e e q n$ $a + b + c = 12$ $p l s c h e c k . . . w h e t h e r a n y d i r e c t m e t h o d t o s o l v e$ $i d o n o t k n o w . . .$
	Commented bymath2018 last updated on 13/Jul/18

	$T h a n k y o u f o r y o u r w o r k!$
	Answered by ajfour last updated on 12/Jul/18

	$l e t a = 3 \tan α, b = 4 \tan β, c = 5 \tan γ$ $T h e n$ $\frac{8 \times 9 \tan^{2} α}{9 \sec^{2} α} = 4 \tan β$ $\Rightarrow 2 \sin^{2} α = \tan β . . . . (i)$ $\frac{10 \times 16 \tan^{2} β}{16 \sec^{2} β} = 5 \tan γ$ $\Rightarrow 2 \sin^{2} β = \tan γ . . . . (i i)$ $\frac{6 \times 25 \tan^{2} γ}{25 \sec^{2} γ} = 3 \tan α$ $\Rightarrow 2 \sin^{2} γ = \tan α . . . . (i i i)$ $(i) \times (i i) \times (i i i) g i v e s$ $8 \sin^{2} α \sin^{2} β \sin^{2} γ = \tan α \tan β \tan γ$ $o r \sin 2 α \sin 2 β \sin 2 γ = 1$ $\Rightarrow α = β = γ = \frac{π}{4}$ $a + b + c = 3 \tan α + 4 \tan β + 5 \tan γ$ $= (3 + 4 + 5) \tan \frac{π}{4}$ $a + b + c = 12 .$ $(w h y t h e m i n i m u m ?)!$
	Commented bytanmay.chaudhury50@gmail.com last updated on 13/Jul/18

	$y e s i f o r g o t i t . . . t h a n k y o u f o r c o r r e c t i o n$
	Commented bymath2018 last updated on 13/Jul/18

	$∵ a > 0, b > 0, c > 0$ $∴ a + b + c \neq 0$ $T h a n k y o u S i r .$
	Commented bymath2018 last updated on 13/Jul/18

	$a = \frac{6 c^{2}}{c^{2} + 25} ⩽ \frac{6 c^{2}}{2 \sqrt{25 c^{2}}} = \frac{3 c}{5}$ $b = \frac{8 a^{2}}{a^{2} + 9} ⩽ \frac{8 a^{2}}{2 \sqrt{9 a^{2}}} = \frac{4 a}{3}$ $c = \frac{10 b^{2}}{b^{2} + 16} ⩽ \frac{10 b^{2}}{2 \sqrt{16 b^{2}}} = \frac{5 b}{4}$ $n o w w e c a n g e t t h e m i n i m u m v a l u e o f a + b + c$ $T h a n k y o u v e r y m u c h!$
	Commented bytanmay.chaudhury50@gmail.com last updated on 13/Jul/18

	$\frac{a}{3} = \frac{b}{4} = \frac{c}{5} = k$ $a = 3 k b = 4 k c = 5 k$ $a = \frac{6 c^{2}}{c^{2} + 25} = \frac{6 \times 25 k^{2}}{25 k^{2} + 25}$ $3 k = \frac{6 k^{2}}{k^{2} + 1}$ $3 k (k^{2} + 1) - 6 k^{2} = 0$ $3 k (k^{2} + 1 - 2 k) = 0$ $s o k = 0 a n d k = 1$ $a + b + c$ $= 3 k + 4 k + 5 k = 12 k$ $s o w h e n k = 0 a + b + c = 0$ $w h e n k = 1 a + b + c = 12$ $h e n c e m i n v a l u e i s z e r o . . . p l s c h e c k . . .$
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