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The following question was a homework and I did not knew how to solve that So I asked my teacher he said the minimum value is achieved at $x = y =z$. the question is as follows:

If $x$, $y$, $z$ are distinct positive real numbers, the minimum value of $$\frac{x^2 (y + z) + y^2 (x + z) + z^2 (x + y)}{xyz}$$

I do not get it why is the solution at x = y = z . the question states "3 distinct real positive numbers" but x = y = z are not distinct numbers.

Also if the question was :

The sum of $2$ distinct positive real numbers is $6$, what is the maximum value of their product?

Would it be a number less than $9$ but not equal to $9$ and what would that number be ?

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  • $\begingroup$ I think that I agree with your interpretation. Usually, distinct means that the variables have all different values. Potentially, the question is stating that you have three variables. $\endgroup$ – Michael Burr May 14 '18 at 20:17
  • $\begingroup$ For the variant you propose (with two numbers), there is no maximum number – only a least upper bound. $\endgroup$ – Bernard May 14 '18 at 20:32
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You are correct that distinct should rule out $x=y=z$ and in your second question it should rule out both numbers being $3$. That makes a problem for both questions as there is no minimum in the first nor maximum in the second. The $\inf$ of the values in the first is $6$ and the $\sup$ is $9$ in the second, but to be a minimum/maximum the value must be achieved.

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By $AM-GM$ we get$$x^2(y+z)+y^2(x+z)+z^2(x+y)\geq 2x^2\sqrt{yz}+2y^2\sqrt{xz}+2z^2\sqrt{xy}\geq 3\sqrt[3]{8x^2y^2z^2xyz}=6xyz$$ Can you get it from here?

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  • $\begingroup$ I don't think this was OP's question. I think OP was focused on the word distinct which disallows $x=y=z$ $\endgroup$ – Ross Millikan May 14 '18 at 20:22
  • $\begingroup$ Yes, I can get it from here but my main question was whether x = y = z is a valid answer or not . Thank you. $\endgroup$ – james May 14 '18 at 20:24
  • $\begingroup$ Ok the minimum will be achieved if $$x=y=z$$ sorry for my confusion $\endgroup$ – Dr. Sonnhard Graubner May 14 '18 at 20:25
  • $\begingroup$ This is a well-known theorem about the AM-GM inequality. $\endgroup$ – Dr. Sonnhard Graubner May 14 '18 at 20:26
  • $\begingroup$ but the question states "distinct" positive real numbers $\endgroup$ – james May 14 '18 at 20:28

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