3.3 Multivariable Differential and Integral Calculus 173Solution.We will show that the best choice forkis^14. To prove this fact, note that the
inequality remains unchanged on replacinga, b, cbyta,tb,tcwitht> 0 .Consequently,
the smallest value ofkis the supremum of
f (a, b, c)=ab
a+b+ 2 c+
bc
b+c+ 2 a+
ca
c+a+ 2 bover the domain ={(a,b,c)|a, b, c > 0 ,a+b+c= 1 }.Note that on ,
f (a, b, c)=ab
1 +c+
bc
1 +a+
ca
1 +b.
To find the maximum of this function on , we will apply the method of Lagrange
multipliers with the constraintg(a, b, c)=a+b+c=1. This yields the system of
equations
b
1 +c+
c
1 +b−
bc
( 1 +a)^2=λ,
c
1 +a+
a
1 +c−
ca
( 1 +b)^2=λ,
a
1 +b+
b
1 +a−
ab
( 1 +c)^2
=λ,a+b+c= 1.Subtracting the first two equations, we obtain
b−a
1 +c+
c
1 +b[
1 +
a
1 +b]
−
c
1 +a[
1 +
b
1 +a]
= 0 ,
which after some algebraic manipulations transforms into
(b−a)[
1
1 +c+
c(a+b+ 1 )(a+b+ 2 )
( 1 +a)^2 ( 1 +b)^2]
= 0.
The second factor is positive, so this equality can hold only ifa=b. Similarly, we prove
thatb=c. So the only extremum offwhen restricted to the planea+b+c=1is
f(
1
3
,
1
3
,
1
3
)
=
1
4
.
But is this amaximum? Let us examine the behavior offon the boundary of (to which
it can be extended). If sayc=0, thenf(a,b, 0 )=ab. Whena+b=1, the maximum
of this expression is again^14. We conclude that the maximum on is indeed^14 , which is
the desired constant.