Given $a,b,c,k>0$ and Maximize $f(x)=x^ay^bz^c$ for $x,y,z \in [0, \infty)$ on $A=\{\,(x,y,z)\mid x^k+y^k+z^k=1\,\}$ 
Given $a,b,c,k>0$ Maximize $f(x)=x^ay^bz^c$ for $x,y,z \in [0, \infty)$ on $A=\{\,(x,y,z)\mid x^k+y^k+z^k=1\,\}$

So I used $g(x,y,z)=x^k+y^k+z^k-1$ with the Lagrange method and obtained through some manipulation:
$f(x)= \lambda k/(a+b+c)$ but I can't really find what is $\lambda$.
 A: We consider $$\Lambda(x,y,z)=x^ay^bz^c-\lambda(x^k+y^k+z^k-1) $$
and by applying $\frac\partial{\partial x}$, $\frac\partial{\partial y}$, $\frac\partial{\partial z}$ are led to the equations
$$\begin{align}ax^{a-1}y^bz^c-k\lambda x^{k-1}&=0\\
bx^ay^{b-1}z^c-k\lambda y^{k-1}&=0\\
cx^ay^{b}z^{c-1}-k\lambda z^{k-1}&=0\\
\end{align}$$
If we add $x\cdot(i)+y\cdot(ii)+z\cdot (iii)$ we obtain your result
$$ (a+b+c)\,\underbrace{x^ay^bz^c}_{f(x,y,z)}-k\lambda=0$$
which is interesting, but not (immediately) helpful (especially as $\lambda$ is the only value we are not interested in).
But substituting this back to eliminate $k\lambda$ from it is helpful: We obtain from the first equation
$$ax^{a-1}y^bz^c-(a+b+c) x^{a+k-1}y^bz^c=0$$
and so (provided $x,y,z\ne 0$, which we may assume as the maximum of $f$ will certainly not be zero)
$$x=\sqrt[k]{\frac a{a+b+c}}. $$
Accordingly, we also find
$$y=\sqrt[k]{\frac b{a+b+c}},\quad z=\sqrt[k]{\frac c{a+b+c}}. $$
So finally the maximum value of $f$ under the given conditions is
$$\sqrt[k]{\frac{a^{a}b^{b}c^{c}}{(a+b+c)^{a+b+c}}}. $$
