If-x-y-z-1-with-0-lt-x-y-z-lt-1-2-then-find-tbe-range-of-values-of-1-x-y-1-y-z-1-z-x-

Question Number 15721 by ajfour last updated on 13/Jun/17

I f \boldsymbol x + \boldsymbol y + \boldsymbol z = 1

w i t h 0 < x, y, z < \frac{1}{2}

t h e n f i n d t b e r a n g e o f v a l u e s o f

\frac{1}{\boldsymbol x + \boldsymbol y} + \frac{1}{\boldsymbol y + \boldsymbol z} + \frac{1}{\boldsymbol z + \boldsymbol x} .

Answered by mrW1 last updated on 13/Jun/17

\frac{1}{\boldsymbol x + \boldsymbol y} + \frac{1}{\boldsymbol y + \boldsymbol z} + \frac{1}{\boldsymbol z + \boldsymbol x} = \frac{1}{x + y} + \frac{1}{1 - x} + \frac{1}{1 - y}

let f (x, y) = \frac{1}{x + y} + \frac{1}{1 - x} + \frac{1}{1 - y}

\frac{\partial f}{\partial x} = - \frac{1}{{(x + y)}^{2}} + \frac{1}{{(1 - x)}^{2}}

\frac{\partial^{2} f}{\partial x^{2}} = \frac{2}{{(x + y)}^{3}} + \frac{2}{{(1 - x)}^{3}}

\frac{\partial f}{\partial y} = - \frac{1}{{(x + y)}^{2}} + \frac{1}{{(1 - y)}^{2}}

\frac{\partial^{2} f}{\partial y^{2}} = \frac{2}{{(x + y)}^{3}} + \frac{2}{{(1 - y)}^{3}}

\frac{\partial^{2} f}{\partial x \partial y} = \frac{2}{{(x + y)}^{3}}

\frac{\partial f}{\partial x} = - \frac{1}{{(x + y)}^{2}} + \frac{1}{{(1 - x)}^{2}} = 0

\Rightarrow x + y = 1 - x \dots (i)

\frac{\partial f}{\partial y} = - \frac{1}{{(x + y)}^{2}} + \frac{1}{{(1 - y)}^{2}} = 0

\Rightarrow x + y = 1 - y \dots (ii)

from (i) and (ii)

\Rightarrow x = y = \frac{1}{3}

f (\frac{1}{3}, \frac{1}{3}) = \frac{1}{\frac{1}{3} + \frac{1}{3}} + \frac{1}{1 - \frac{1}{3}} + \frac{1}{1 - \frac{1}{3}} = \frac{9}{2}

\frac{\partial^{2} f (\frac{1}{3}, \frac{1}{3})}{\partial x^{2}} = \frac{2}{{(\frac{2}{3})}^{3}} + \frac{2}{{(\frac{2}{3})}^{3}} = \frac{27}{2} > 0

\frac{\partial^{2} f (\frac{1}{3}, \frac{1}{3})}{\partial y^{2}} = \frac{2}{{(\frac{2}{3})}^{3}} + \frac{2}{{(\frac{2}{3})}^{3}} = \frac{27}{2} > 0

\frac{\partial^{2} f (\frac{1}{3}, \frac{1}{3})}{\partial x \partial y} = \frac{2}{{(\frac{2}{3})}^{3}} = \frac{27}{4} > 0

\Rightarrow f (\frac{1}{3}, \frac{1}{3}) = \frac{9}{2} is the minimum!

\underset{x \to 0, y \to 0}{\lim f} (x, y)

= \lim_{x \to 0} [\underset{y \to 0}{\lim f} (x, y)]

= \lim_{x \to 0} [\lim_{y \to 0} (\frac{1}{x + y} + \frac{1}{1 - x} + \frac{1}{1 - y})]

= \lim_{x \to 0} [\frac{1}{x} + \frac{1}{1 - x} + 1]

= \lim_{x \to 0} [\frac{1}{x}] + 1 + 1

= + \infty

\Rightarrow \frac{1}{\boldsymbol x + \boldsymbol y} + \frac{1}{\boldsymbol y + \boldsymbol z} + \frac{1}{\boldsymbol z + \boldsymbol x} \in [\frac{9}{2}, + \infty)

Commented by ajfour last updated on 13/Jun/17

t h a n k s s i r, y o u a r e f a r t o o

g e n e r o u s!

Commented by mrW1 last updated on 13/Jun/17

thank you sir! i do also learn from you .

thank you sir! i do also learn from you .

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