What′s the minimum value of ((13a+13b+2c)/(2a+2b))+((24a−b+13c)/(2b+2c))+((−a+24b+13c)/(2a+2c))? (a,b,c are positive numbers.) I think nobody can solve this.


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Question Number 76061 by tw000001 last updated on 23/Dec/19

${What}^{'} s the minimum value of$ $\frac{13 a + 13 b + 2 c}{2 a + 2 b} + \frac{24 a - b + 13 c}{2 b + 2 c} + \frac{- a + 24 b + 13 c}{2 a + 2 c} ?$ $(a, b, c are positive numbers .)$ $I think nobody can solve this .$
	Answered by MJS last updated on 23/Dec/19

	$\frac{13 a + 13 b + 2 c}{2 a + 2 b} + \frac{24 a - b + 13 c}{2 b + 2 c} + \frac{- a + 24 b + 13 c}{2 a + 2 c} =$ $= \frac{c}{a + b} + \frac{12 b + 7 c}{a + c} + \frac{12 a + 7 c}{b + c} + \frac{11}{2}$ $we see that a and b are of equal weight but$ $c is different$ $let a = p c \land b = q c$ $\frac{1}{p + q} + \frac{12 q + 7}{p + 1} + \frac{12 p + 7}{q + 1} + \frac{11}{2}$ $to ensure positive values$ $let p = x^{2} \land q = y^{2}$ $\frac{1}{x^{2} + y^{2}} + \frac{12 y^{2} + 7}{x^{2} + 1} + \frac{12 x^{2} + 7}{y^{2} + 1} + \frac{11}{2}$ $now I simply tried y = x$ $- \frac{10}{x^{2} + 1} + \frac{1}{2 x^{2}} + \frac{59}{2} = \frac{59 x^{4} + 40 x^{2} + 1}{2 x^{2} (x^{2} + 1)}$ $\frac{d}{d x} [\frac{59 x^{4} + 40 x^{2} + 1}{2 x^{2} (x^{2} + 1)}] = 0$ $\frac{19 x^{4} - 2 x^{2} - 1}{x^{3} {(x^{2} + 1)}^{2}} = 0 \Rightarrow x^{2} = \frac{1 + 2 \sqrt{5}}{19}$ $\Rightarrow \min \frac{59 x^{4} + 40 x^{2} + 1}{2 x^{2} (x^{2} + 1)} = 19 + 2 \sqrt{5} \approx 23.472136$ $at a = b = \frac{1 + 2 \sqrt{5}}{19} c; c \in R^{+}$
	Commented by MJS last updated on 23/Dec/19
	btw my name is Nobody ��
	Commented by peter frank last updated on 23/Dec/19

	$h a h a h a h a h g r e a t m a n m r m j s$
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