+179 Let x, y and z be positive real numbers such that x + y + z = 1.
Find the minimum value of \(\frac{x + y}{xyz}\).
After submitting my first answer, I found that the minimum value does not occur when x=y=z=1/3, meaning that the answer must be less than 18 (because 18 is achievable).
Thanks for any help, especially if you could briefly describe the proof you used.
Have a great day!
+2407 Your questions are always so hard but I'll try my best.
Since x and y seem to be used the same in the equation.
I'm going to assume that x = y. *This could be totally incorect.
(2x)/(x^2z) is our new equation.
z = 1 - 2x
Let a be (x+y)/(xyz).
(2x)/(x^2(1-2x)) = a
2x = a(x^2 - 2x^3)
Just quickly graphing this out in desmos by setting a = y (not the y in the original equation, but for coordinate purposes), it seems like the sum can be any value other than from the range (0, 16].
Not sure what went wrong, but I hope this helped a bit.
Good luck.
=^._.^=
+179 Ay! Turns out, the answer was 16, occurring when x=y=1/4 and z=1/2.
We use a combination of the AM-HM and the AM-GM in that order.
Starting, we have by the AM-HM \(\frac{x + y}{2} \ge \frac{2}{\frac{1}{x} + \frac{1}{y}} = \frac{2xy}{x + y}\)
So we have \(\frac{x + y}{xy} \ge \frac{4}{x + y}\).
Multiplying by 1/z, we get \(\frac{x + y}{xyz} \ge \frac{4}{(x + y)z}\).
By the AM-GM inequality,
\(\sqrt{(x + y)z} \le \frac{x + y + z}{2} = \frac{1}{2}\),
meaning \((x + y)z \le \frac{1}{4}\).
Hence,
\(\frac{4}{(x + y)z} \ge 16\).
Using what we had before, we can write
\(\frac{x + y}{xy} \ge \frac{4}{x + y} \ge 16\).
Equality occurs when x=y, and the minimum value is 16.
If you ask me, this solution was quite interesting. I thought of using the QM-AM-GM-HM relationship, as well as the Cauchy-Schwarz Inequality, but not in this way. Thanks for your help!
