Question 1143014
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By analogy with the well known AM-GM inequality ("Arithmetic Mean - Geometric Mean inequality") for two variables "a" and "b"


    ab <= {{{((a+b)/2)^2}}},         (1)


there is AM-GM inequality for three variables "a", "b" and "c"


    abc <= {{{((a + b + c)/3)^3}}}.      (2)


Inequalities (1) and (2) are valid for any two and three variables, respectively, that are real non-negative numbers.


Apply inequality (2), taking  


    a = x-3,  b = y-5,  c = z-2.


You will get


    (x-3)*(y-5)*(z-2) <= {{{(((x-3)+(y-5)+(z-2))/3)^3}}} = {{{((x+y+z - 3-5-2)/3)^3}}} = {{{((16-10)/3)^3}}} = {{{(6/3)^3}}} = {{{2^3}}} = 8.


Thus for any 3 values of x, y and z, restricted by the equality


    x + y + z = 16   and  inequalities  x >= 3, y >= 5  and  z >= 2,


the inequality  


    (x-3)*(y-5)*(z-2) <= 8 


is held.


From the other side, at x= 5, y= 7  and z= 4 we have


    (x-3)*(y-5)*(z-2) = (5-3)*(7-5)*(4-2) = 2*2*2 = 8.


and the values of x, y and z satisfy all needed restrictions.



Thus the maximum value of (x-3)*(y-5)*(z-2), where  x, y and z are restricted by 


    x + y + z = 16,    x >= 3, y >= 5  and  z >= 2


is 8.        <U>ANSWER</U>
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