Solve #a^2+b^2+c^2>= a+b+c# ?

1 Answer
Feb 1, 2017

Answer:

See below.

Explanation:

#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