Find the greatest value of #abc# for positive values of #a,b,c# subject to #ab +bc+ca=12# ?

Here's a method using a Lagrangian...

Let:

#{ (a=e^u), (b=e^v), (c=e^w) :}#

where #u, v, w# are real numbers.

That takes care of the requirement that #a, b, c > 0#.

We want to maximise:

#f(u,v,w) = e^ue^ve^w#

subject to:

#g(u,v,w) = e^ue^v+e^ve^w+e^we^u = 12#

Define the Lagrangian:

#L(u,v,w,lambda) = f(u,v,w)-lambda(g(u,v,w)-12)#

#color(white)(L(u,v,w,lambda)) = e^ue^ve^w-lambda(e^ue^v+e^ve^w+e^we^u-12)#

At the maxima and minima we have:

#grad L(u,v,w,lambda) = bb(0)#

That is, all of the partial derivatives of #L# are zero:

#0 = (del L)/(del u) = e^u(e^ve^w-lambda (e^v+e^w))#

#0 = (del L)/(del v) = e^v(e^we^u-lambda (e^w+e^u))#

#0 = (del L)/(del w) = e^w(e^ue^v-lambda (e^u+e^v))#

#0 = (del L)/(del lambda) = -(e^ue^v+e^ve^w+e^we^u-12)#

That is:

#{ (bc-lambda(b+c) = 0), (ca-lambda(c+a) = 0), (ab-lambda(a+b)=0), (ab+bc+ca=12) :}#

We find:

#lambda = (bc)/(b+c) = (ca)/(c+a) = (ab)/(a+b)#

Hence:

#c = (lambdab)/(b-lambda) = (lambdaa)/(a-lambda)#

Hence:

#lambdaab-lambda^2b = lambdaab-lambda^2a#

Hence #a=b# and similarly #b=c#

Hence #a=b=c# and #12 = 3a^2#, so #a=2# and #abc=8#

#color(white)()#
Footnote

The Lagrangian method used here is a standard method for finding maxima and minima of continuous functions of several variables subject to constraints.

In this particular example, there may be an alternative algebraic method involving #(a-b)^2+(b-c)^2+(c-a)^2# or similar, especially since we have found that the solution occurs when #a=b=c#.