I need to show absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ is $\frac{a^ab^bc^c}{(a+b+c)^{a+b+c}}$
So I I do have $ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$
then I went on to equating each of the 2 equations giving
$y=\frac{b}{a}x, z=\frac{c}{a}x$
So I have $x+\frac{b}{a}x+\frac{c}{a}x=1$ but I am not so sure how to continue
The answer instead had
$\lambda x = ax^ay^bz^c, \lambda y = bx^ay^bz^c, \lambda z = cx^ay^bz^c$, then making observation that $x:y:z=a:b:c$, which means
$x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$
I don't quite get this ... how do I derive this?
The rest ...
$(\frac{a}{a+b+c})^a(\frac{b}{a+b+c})^b(\frac{c}{a+b+c})^c=1$
...