SOLUTION: (a) Let x, y, and z be positive real numbers. Find the largest possible value of \sqrt{\frac{3x + 5y + 2z}{6x + 5y + 4z}} + \sqrt{\frac{2x + 5y + z}{6x + 5y + 5z}} + \sqrt{\frac{

Algebra ->  Expressions-with-variables -> SOLUTION: (a) Let x, y, and z be positive real numbers. Find the largest possible value of \sqrt{\frac{3x + 5y + 2z}{6x + 5y + 4z}} + \sqrt{\frac{2x + 5y + z}{6x + 5y + 5z}} + \sqrt{\frac{      Log On


   



Question 1209926: (a) Let x, y, and z be positive real numbers. Find the largest possible value of
\sqrt{\frac{3x + 5y + 2z}{6x + 5y + 4z}} + \sqrt{\frac{2x + 5y + z}{6x + 5y + 5z}} + \sqrt{\frac{9x + y + 4z}{6x + 5y + 4z}}.

(b) Find \frac{z}{x} if (x,y,z) is a triple that gives the maximum value in Part (a).

Found 3 solutions by Edwin McCravy, ikleyn, mccravyedwin:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
That crazy notation again!  Why they insist on that, I'll neve know.

(a) Let x, y, and z be positive real numbers.  Find the largest possible value of


The arithmetic mean of expressions for non-negative numbers always exceeds the
geometric mean except when they are all the same.  So we normalize them by
setting x=y=z= a constant, which may as well be 1.  So we let x=y=z=1 and

sqrt%2810%2F15%29%2Bsqrt%288%2F16%29%2Bsqrt%2814%2F15%29%22%22=%22%22

sqrt%282%2F3%29%2Bsqrt%281%2F2%29%2Bsqrt%2814%2F15%29

That's the largest possible value.

(b) Find z%2Fx if (x,y,z) is a triple that gives the maximum value in Part
(a).

Since (1,1,1) is such a triple z%2Fx=1%2F1=1.  No?  

Edwin

Answer by ikleyn(52767) About Me  (Show Source):
You can put this solution on YOUR website!
.
(a) Let x, y, and z be positive real numbers. Find the largest possible value of
\sqrt{\frac{3x + 5y + 2z}{6x + 5y + 4z}} + \sqrt{\frac{2x + 5y + z}{6x + 5y + 5z}} + \sqrt{\frac{9x + y + 4z}{6x + 5y + 4z}}.
(b) Find \frac{z}{x} if (x,y,z) is a triple that gives the maximum value in Part (a).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


The solution in the post by  Edwin  (as  Edwin  McCravy)  is incorrect.

It becomes incorrect starting from Edwin's assumption that x = y = z provide the maximum.

Edwin refers to the fact that "the arithmetic mean of expressions for non-negative numbers
always exceeds the geometric mean except when they are all the same".

This fact is true, but it has nothing in common with the assumption x = y = z.

This fact only means that the three expressions under the square roots should be equal,
but this does not implies or is equivalent to x = y = z.


\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\


In his post as @mccravyedwin, Edwin writes 

    "It doesn't seem obvious to Ikleyn why it works just to substitute x = y = z = 1."


In response, I will say that 

    It doesn't seem obvious to  highlight%28highlight%28Edwin%29%29  why it  highlight%28highlight%28DOES%29%29  highlight%28highlight%28NOT%29%29  work just to substitute x = y = z = 1.


To prove it, I will take x = y = 1, z = 0.  In other words, I will assume that x and y both go to infinity
with the same rate, while z remains limited.


Then the expression 




will approach to 

 =  = 2.60398949 approximately, which is  highlight%28highlight%28greater%29%29  than Edwin's number 2.489695145.


                So, Edwin's invectives to my address have no any solid base.



Answer by mccravyedwin(406) About Me  (Show Source):
You can put this solution on YOUR website!

It doesn't seem obvious to Ikleyn why it works just to substitute x = y = z = 1.
Let's forget the means of x,y,z,w.  I was thinking of that to get a starting
place to start my thinking from.

(a) Let x, y, and z be positive real numbers.  Find the largest possible value of


To me, it seems obvious that if all three variables approach infinity at the
same rate, they will approach 

%22%22=%22%22%22%22=%22%22sqrt%282%2F3%29%2Bsqrt%281%2F2%29%2Bsqrt%2814%2F15%29%22%22=%22%22
2.489695145 approximately, and that the value could not be higher than that,

Notice that is the same value as substituting 1 for all the variables.  I am not
saying that proves anything, just that it starts us to thinking in the right
direction.  I still think it's right.  Maybe we could show it rigorously with
multivariable calculus by setting the partial derivatives equal zero, and
examining the behavior at that point.

(b) Find z%2Fx if (x,y,z) is a triple that gives the maximum value in Part
(a).

Since (1,1,1) is such a triple z%2Fx=1%2F1=1.   

Edwin