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I\! f x, y>0, then\; the\; minimum\; value\; o\! f\\

(x+y)(y+z)\left ( \frac{1}{x}+\frac{1}{y} \right )\left ( \frac{1}{y}+\frac{1}{z} \right )

Option: 1

4


Option: 2

8


Option: 3

16


Option: 4

None of these


Answers (1)

best_answer

\\Apply \;AM\geqslant GM\;on \; x,y\\\\ \frac{x+y}{2}\geqslant \sqrt{xy}

\\\Rightarrow(x+y) \geqslant 2 \sqrt{x y} \quad \text {..... (i) }\\\\ Apply \,\,A M \geqslant GM\; on\; \frac{1}{x}, \frac{1}{y}. \\\frac{1}{x}+\frac{1}{y} \geqslant 2 \sqrt{\frac{1}{x y}}\,\,\,\,\,\,\,\,....(ii)\\\\ (i)\times(i i)\\ (x+y)\left(\frac{1}{x}+\frac{1}{y}\right) \geqslant 4\,\,\,\,\,\,\,\,....(iii)\\\\ Similarly \\ (y+z)\left(\frac{1}{y}+\frac{1}{z}\right) \geqslant 4\quad\,\,\,\,\,\,\,\,....(i v)\\\\(i i i) \times (i v) \\(x+y)(y+z)\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right) \geqslant 16

So minimum value is 16

Posted by

shivangi.shekhar

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