Given-a-gt-0-b-gt-0-c-gt-2-a-b-1-find-the-minimum-value-of-3ac-b-c-ab-6-c-2-

Question Number 179090 by CrispyXYZ last updated on 24/Oct/22

Given a > 0, b > 0, c > 2, a + b = 1 .

find the minimum value of \frac{3 a c}{b} + \frac{c}{a b} + \frac{6}{c - 2} .

Commented by Frix last updated on 24/Oct/22

24

Answered by MJS_new last updated on 24/Oct/22

a > 0 \land b > 0 \land a + b = 1 \Rightarrow 0 < a < 1 \land 0 < b < 1

\Rightarrow let a = \sin^{2} x \land b = \cos^{2} x

I prefer x = \arctan t \Rightarrow a = \frac{t^{2}}{t^{2} + 1} \land b = \frac{1}{t^{2} + 1}

c > 2 \Rightarrow c = 2 + u^{2}

\Rightarrow

\frac{3 a c}{b} + \frac{c}{a b} + \frac{6}{c - 2} =

= 3 t^{2} (u^{2} + 2) + \frac{{(t^{2} + 1)}^{2} (u^{2} + 2)}{t^{2}} + \frac{6}{u^{2}} = f (t, u)

\frac{d f}{d t} = 0

6 (u^{2} + 2) t + \frac{2 (t^{4} - 1) (u^{2} + 2)}{t^{3}} = 0

2 (4 t^{4} - 1) (u^{2} + 2) = 0

t = \pm \frac{\sqrt{2}}{2}

f (\pm \frac{\sqrt{2}}{2}, u) = \frac{3 (u^{2} + 2)}{2} + \frac{9 (u^{2} + 2)}{2} + \frac{6}{u^{2}} = \frac{6 (u^{2} + 1)}{u^{2}}

\frac{d f}{d u} = 0

\frac{12 (u^{4} - 1)}{u^{3}} = 0

u = \pm 1

f (\pm \frac{\sqrt{2}}{2}, \pm 1) = 24

with a = \frac{1}{3} \land b = \frac{2}{3} \land c = 3

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