Question 461623
The first problem can be solved by the AM-GM inequality.  I will solve the second problem using calculus.

{{{1/3 = (x+2y + 3z)/3 >= root(3, x*2y*3z) = root(3, 6xyz)}}}

==> {{{1/3 >= root(3, 6xyz)}}} ==> {{{1/27 >= 6xyz}}}  (because the function y = x^3 is 1-to-1.)

==> {{{1/162 >= xyz}}}.

Hence the maximum value of xyz is 1/162, and it happens if and only if x = y = z = {{{(1/3)*root(3, 1/6)}}}.

For the second problem, 

v = 1- u, so that {{{u^2v = u^2(1 - u) = u^2 - u^3}}}

==> {{{d(u^2 - u^3)/du = 2u - 3u^2 = 0}}} ==> u = 0, 2/3.  There is no need to test for u = 0.

{{{d^2(u^2 - u^3)/du^2 = 2 - 6u = 2 - 6*(2/3) = -2 <0}}} when u = 2/3.
==> there is absolute max when u = 2/3, by the 2nd derivative test.

==> the max value of {{{u^2v}}} is {{{(2/3)^2*(1/3) = 4/27}}}.