abc-64-a-b-c-R-Find-K-that-satisfy-to-the-inequality-a-b-ab-b-c-bc-c-a-ca-abc-a-b-c-K-

Question Number 59273 by naka3546 last updated on 07/May/19

a b c = 64

a, b, c \in R^{+}

F i n d K t h a t s a t i s f y t o t h e i n e q u a l i t y :

\frac{(a + b) \sqrt{a b} + (b + c) \sqrt{b c} + (c + a) \sqrt{c a}}{\sqrt{a b c}} ⩾ \sqrt{a} + \sqrt{b} + \sqrt{c} + K .

Answered by tanmay last updated on 07/May/19

\frac{a + b}{2} ⩾ \sqrt{a b}

(a + b) ⩾ 2 \sqrt{a b}

(a + b) \sqrt{a b} ⩾ 2 a b

s o (a + b) \sqrt{a b} + (b + c) \sqrt{b c} + (c + a) \sqrt{a c} ⩾ 2 (a b + b c + c a)

L H S

\frac{N_{r}}{D_{r}} ⩾ \frac{2 (a b + b c + c a)}{\sqrt{a b c}}

\frac{N_{r}}{D_{r}} ⩾ \frac{a b + b c + c a}{4}

\frac{a + b + c}{3} ⩾ {(a b c)}^{\frac{1}{3}}

(a + b + c) ⩾ 12

★ n o w \frac{a b + b c + c a}{3} ⩾ {(a b \times b c \times c a)}^{\frac{1}{3}}

\frac{a b + b c + c a}{3} ⩾ {(a b c)}^{\frac{2}{3}}

\frac{a b + b c + c a}{4} ⩾ \frac{3}{4} \times {(64)}^{\frac{2}{3}}

\frac{a b + b c + c a}{4} ⩾ \frac{3}{4} \times 16

\frac{a b + b c + c a}{4} ⩾ 12 ★

\frac{N_{r}}{D_{r}} ⩾ \frac{a b + b c + c a}{4}

\frac{N_{r}}{D_{r}} ⩾ \frac{a b + b c + c a}{4} ⩾ 12

R H S

\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{k}

\frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{3} ⩾ {(\sqrt{a} \times \sqrt{b} \times \sqrt{c})}^{\frac{1}{3}}

\frac{\sqrt{a} + \sqrt{b} + \sqrt{c}}{3} ⩾ {(a b c)}^{\frac{1}{6}}

(\sqrt{a} + \sqrt{b} + \sqrt{c}) / 3 ⩾ {(64)}^{\frac{1}{6}} \leftarrow c o r r e c t i o n a f t e r N a k a s i r d e t e c t e d

\sqrt{a} + \sqrt{b} + \sqrt{c} ⩾ 2 \times 3 = 6

s o

\frac{N_{r}}{D_{r}} ⩾ \frac{a b + b c + c a}{4} ⩾ 12 ⩾ 6 + k

s o w h e n 12 = 6 + k k = 6

w h e n 12 ⩾ 6 + k

6 ⩾ k

i h a v e j u s t t r i e d \dots

Commented by naka3546 last updated on 07/May/19

\sqrt{a} + \sqrt{b} + \sqrt{c} ⩾ 3 \cdot 2 = 6 ?

\frac{N_{r}}{D_{r}} ⩾ \frac{a b + b c + c a}{4} ⩾ 12 ⩾ 2 + k

s o w h e n 12 = 6 + k k = 6

K = 6

Commented by tanmay last updated on 07/May/19

t h a n k y o u s i r f o r y o u r k i n d p e r u s a l a n d d e t e c t i o n

o f e r r o r \dots

Facebook Tweet Pin