Absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ 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 ... 
$$(\frac{a}{a+b+c})^a(\frac{b}{a+b+c})^b(\frac{c}{a+b+c})^c=1$$ 
...
 A: You were doing fine, the argument from the textbook that you are quoting is more symmetrical, that's all.
You obtained $x+\frac{b}{a}x+\frac{c}{a}x=1$. That almost finishes things! Multiply through by $a$. You get $(a+b+c)x=a$ and therefore 
$$x=\frac{a}{a+b+c}.$$
From your $y=\frac{b}{a}x$ you can then get $y=\dfrac{b}{a+b+c}$, and similarly $z=\dfrac{c}{a+b+c}$.  That gets you to exactly the same place as the solution you quoted.
A: When you have
$$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$
$ax^{a-1}y^bz^c =\lambda \hspace{4pt}$ will give you $\hspace{4pt} \lambda x = a x^a y^b z^c$
Similarly
$bx^ay^{b-1}z^c =\lambda \hspace{4pt} \Rightarrow \hspace{4pt} \lambda y = b x^a y^b z^c$
$cx^ay^bz^{c-1} =\lambda \hspace{4pt} \Rightarrow \hspace{4pt} \lambda z = c x^a y^b z^c$
$$ \frac{\lambda x}{\lambda y} = \frac{x}{y} = \frac{a x^a y^b z^c}{b x^a y^b z^c} = \frac{a}{b}$$
$$ \frac{\lambda y}{\lambda z} = \frac{y}{z} = \frac{b x^a y^b z^c}{cx^a y^b z^c} = \frac{b}{c}$$
$$ x : y : z = a : b : c$$
What happens to $$\lambda x + \lambda y + \lambda z  \hspace{5pt} ?$$
Is it 
$$ x^a y^b z^c (a+b+c)$$ 
Now can you relate to the answer you were supposed to get?
