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-

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