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Question Number 116323 by bobhans last updated on 03/Oct/20

(1) Let a, b and c real number such that

\frac{ab}{a + b} = \frac{1}{3}, \frac{bc}{b + c} = \frac{1}{4} and \frac{ac}{a + c} = \frac{1}{5} . Find

the value of \frac{24 abc}{ab + ac + bc} ?

(2) Let p and q be two real number that

satisfy p . q = 2013 . What is the minimum

value of {(p + q)}^{2} ?

Answered by john santu last updated on 03/Oct/20

(1) \to {\begin{cases} \frac{1}{a} + \frac{1}{b} = 3 \\ \frac{1}{b} + \frac{1}{c} = 4 \\ \frac{1}{a} + \frac{1}{c} = 5 \end{cases}

s u m m i n g t h e t h r e e e q u a t i o n

\Rightarrow 2 (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = 12

\Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 6; \frac{a b + a c + b c}{a b c} = 6

\Rightarrow \frac{a b c}{a b + a c + b c} = \frac{1}{6}, m u l t i p l y b o t h

s i d e s b y 24 w e g e t \frac{24 a b c}{a b + a c + b c} = 4

Answered by john santu last updated on 03/Oct/20

(2) {(p + q)}^{2} = {(p - q)}^{2} + 4 p q ⩾ 0 + 4 \times 2013

⩾ 8052, w h e n p = q = \sqrt{2013}

M i n i m u m v a l u e o f {(p + q)}^{2} e q u a l t o 8052