#a^2+b^2+c^2>= a+b+c->a^2+b^2+c^2-a-b-c>= 0#
So we can arrange this as an optimization problem
Find
#(a,b,c)_@= "argmin"(a^2+b^2+c^2-a-b-c)#
subjected to #abc=1#
If this solution gives
#a_@^2+b_@^2+c_@^2-a_@-b_@-c_@ ge 0# then the question is proved.
To prove this we will use the Lagrange Multipliers technique. The lagrangian is
#L(a,b,c,lambda)=a^2+b^2+c^2-a-b-c+lambda(abc-1)#
The stationary point conditions are
#{(2 a-1 + b c lambda=0),( 2 y-1 + a c lambda=0), (2 c-1 +
a b lambda=0),( a b c-1=0):}#
or
#{(2 a-1 + lambda/a=0),( 2 b-1 + lambda/b=0), (2 c-1 +
lambda/c=0),( a b c-1=0):}#
which gives
#a=b=c=1/4 (1 pm sqrt(1 - 8 lambda))#
or
#(1/4 (1 pm sqrt(1 - 8 lambda)))^3=1->1/4 (1 pm sqrt(1 - 8 lambda))=1#
giving #lambda=-1#
The solution is #a_@=b_@=c_@ = 1#
This is a minimum point because the hessian of #a^2+b^2+c^2-a-b-c# is #((2,0,0),(0,2,0),(0,0,2))# which is definite positive.
The minimum value is #0#. So it is proved
