+0  
 
0
48
3
avatar

For positive real numbers x, y   and  z compute the maximum value of

 

 

 

 

 

\(\[\frac{xyz(x + y + z)}{(x + y)^2 (y + z)^2}.\]\)
 

 Jul 18, 2023
 #1
avatar
0

 

We can write the expression as [\frac{xyz(x + y + z)}{(x + y)^2 (y + z)^2} = \frac{xy(z(x + y) + 1)}{(x + y)^2}.]By AM-GM, [xy(z(x + y) + 1) \le \sqrt{xy \cdot xy \cdot (z(x + y) + 1)^2} = \sqrt{(xy)(z(x + y))^2 (1 + 1/z)^2} = \sqrt{(x + y)^2 (y + z)^2},]so [\frac{xy(z(x + y) + 1)}{(x + y)^2} \le 1.]

Equality occurs when z=yx+y​, so the maximum value is 1​.

 Jul 18, 2023
 #3
avatar
0

 

         x   y  z    Maximum Value

1  = (3, 1, 2) = 0.25
2  = (2, 1, 3) = 0.25
3  = (6, 2, 4) = 0.25
4  = (4, 2, 6) = 0.25
5  = (9, 3, 6) = 0.25
6  = (6, 3, 9) = 0.25
 

 Jul 18, 2023

5 Online Users

avatar
avatar
avatar