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avatar+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!

 Jun 30, 2021
 #1
avatar+2104 
+3

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. 

 

=^._.^=

 Jun 30, 2021
 #2
avatar+179 
+4

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! 

EnchantedLava68  Jun 30, 2021
 #3
avatar+2104 
+2

YAYYY. :DDDD

That is such an intresting solution. 

I have not yet learned about AM-GM/AM-HM, but I'm planning to do so soon through alcumus. 

I think the mistake I made was forgetting that x, y, and z were all positive, meaning that (x+y)/(xyz) was also positive. 

 

=^._.^=

catmg  Jul 1, 2021

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