a^2+b^2+c^2>= a+b+c->a^2+b^2+c^2-a-b-c>= 0a2+b2+c2≥a+b+c→a2+b2+c2−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)(a,b,c)∘=argmin(a2+b2+c2−a−b−c)
subjected to abc=1abc=1
If this solution gives
a_@^2+b_@^2+c_@^2-a_@-b_@-c_@ ge 0a2∘+b2∘+c2∘−a∘−b∘−c∘≥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)L(a,b,c,λ)=a2+b2+c2−a−b−c+λ(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
