Question 1019866
The AM-GM inequality for three variables states that {{{(x[1] +x[2]+x[3])/3 >= (x[1]*x[2]*x[3])^(1/3)}}} for non-negative {{{x[1]}}}, {{{x[2]}}}, and {{{x[3]}}}.
Hence {{{(a^2 +a + 1)/3 >= (a^2*a*1)^(1/3) = (a^3)^(1/3) = a}}}, or {{{a^2+a+1>=3a}}}  (assuming a is non-negative).
Making the similar arguments for the variables b and c, 
we get

{{{b^2+b+1>=3b}}} and {{{c^2+c+1>=3c}}}.

Thus, after multiplying corresponding sides of the three inequalities, we get

{{{(a^2+a+1)*(b^2 + b+1)*(c^2+c+1)>=27abc}}}, and the greatest number x is 27.