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Let x, y,z be positive numbers. The least value of

$ \frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{.5}}$ is

a) $\frac{9}{2^{.5}}$

b) 6

c) $\frac{1}{6^{.5}}$

d.) None of the above

I tried applying the A.M. -G.M> inequality to solve this , however couldn't succeed. I am not so sure if that is the way to go forward , if you know of a better approach please answer

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closed as off-topic by Carl Mummert, Claude Leibovici, Namaste, Daniel W. Farlow, user223391 Feb 7 '17 at 22:21

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If $x,y,z>0$, then by AM-GM:

$$\frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{1/2}}=\frac{x+y+z+xy+yz+zx}{(xyz)^{1/2}}$$

$$\ge \frac{6\sqrt[6]{x\cdot y\cdot z\cdot xy\cdot yz\cdot zx}}{(xyz)^{1/2}}=6$$

Equality holds if and only if (iff) $x=y=z=xy=yz=zx$, i.e. iff $x=y=z=1$.

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  • $\begingroup$ Hey very nice answer, I have a further question though is it necessary for any A.M.-G.M. inequality that the equality holds only in the case when all the numbers being considered are equal ? $\endgroup$ – Noob101 Apr 30 '16 at 8:25
  • $\begingroup$ @SuryakantShrivastava Yes. See en.wikipedia.org/wiki/…. If $x_1,x_2,\ldots,x_n>0$, then $x_1+x_2+\cdots+x_n\ge n\sqrt[n]{x_1x_2\cdots x_n}$ and equality holds if and only if $x_1=x_2=\cdots=x_n$ (meaning that for equality to hold we must have and it's sufficient to have $x_1=x_2=\cdots =x_n$). $\endgroup$ – user236182 Apr 30 '16 at 8:28
  • $\begingroup$ How did you get to know that numerator would be minimised if all the terms are equal to 1 $\endgroup$ – Noob101 Apr 30 '16 at 8:31
  • $\begingroup$ @SuryakantShrivastava I directly applied AM-GM to prove that $\frac{x(1+y)+y(1+z)+z(1+x)}{(xyz)^{1/2}}\ge 6$. Then, since the terms I used in the AM-GM inequality are $x,y,z,xy,yz,zx$, equality holds if and only if $x=y=z=xy=yz=zx$, i.e. $x=y=z=1$. $\endgroup$ – user236182 Apr 30 '16 at 8:35
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Hint the least value of AM and max value of-GM is obtained when all numbers are equal so $x=y=z$ so least value is $6$ by putting all as $1$

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