Real Analysis 567The method of Lagrange multipliers gives rise to the system of equations
(λ− 1 )
−a^3 +a(b^2 +c^2 )
4=
a+b+c
2,
(λ− 1 )−b^3 +b(a^2 +c^2 )
4=
a+b+c
2,
(λ− 1 )
−c^3 +c(a^2 +b^2 )
4=
a+b+c
2,
g(a, b, c)= 0.Becausea+b+c =0,λcannot be 1, so this further gives
−a^3 +a(b^2 +c^2 )=−b^3 +b(a^2 +c^2 )=−c^3 +c(a^2 +b^2 ).The first equality can be written as(b−a)(a^2 +b^2 −c^2 )=0. This can happen only if
eithera=bora^2 +b^2 =c^2 , so either the triangle is isosceles, or it is right. Repeating
this for all three pairs of sides we find that eitherb=corb^2 +c^2 =a^2 , and also that
eithera=cora^2 +c^2 =b^2. Since at most one equality of the forma^2 +b^2 =c^2 can
hold, we see that, in fact, all three sides must be equal. So the critical point given by the
method of Lagrange multipliers is the equilateral triangle.
Is this the global minimum? We just need to observe that as the triangle degenerates,
the area becomes infinite. So the answer is yes, the equilateral triangle minimizes the area.
512.Consider the functionf:{(a,b,c,d)|a, b, c, d≥ 1 ,a+b+c+d= 1 }→R,
f (a, b, c, d)=1
27
+
176
27
abcd−abc−bcd−cda−dab.Being a continuous function on a closed and bounded set inR^4 ,fhas a minimum. We
claim that the minimum offis nonnegative. The inequalityf (a, b, c, d)≥0 is easy
on the boundary, for if one of the four numbers is zero, sayd=0, thenf (a, b, c, 0 )=
1
27 −abc, and this is nonnegative by the AM–GM inequality.
Any minimum in the interior of the domain should arise by applying the method of
Lagrange multipliers. This method gives rise to the system
∂f
∂a=
176
27
bcd−bc−cd−db=λ,
∂f
∂b=
176
27
acd−ac−cd−ad=λ,
∂f
∂c=
176
27
abd−ab−ad−bd=λ,
∂f
∂d=
176
27
abc−ab−bc−ac=λ,a+b+c+d= 1.