Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$ 
Given $a,b,c,k > 0$, find the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$

The subject is Lagrange multipliers, thus that is what I tried to use, where the gradients are
$$\triangledown f=\begin{pmatrix}ax^{a-1}y^{b}z^{c}\\
bx^{a}y^{b-1}z^{c}\\
cx^{a}y^{b}z^{c-1}
\end{pmatrix}
 $$
and if $g$ is the function of the constraint, then
$$\triangledown g=\begin{pmatrix}kx^{k-1}\\
ky^{k-1}\\
kz^{k-1}
\end{pmatrix}$$
but then I'm having troubles finding the actual maximum, as I'm interested in $x,y,z,\lambda$ satisfying
$$
\begin{pmatrix}ax^{a-1}y^{b}z^{c}\\
bx^{a}y^{b-1}z^{c}\\
cx^{a}y^{b}z^{c-1}
\end{pmatrix}=\lambda\begin{pmatrix}kx^{k-1}\\
ky^{k-1}\\
kz^{k-1}
\end{pmatrix}
$$
and ignoring for a second the possibilities of one of the constants being $1$, this gives some equations which really don't seem all that right, as extracting $x$ from the first equality for example gives
$$x=\left(\frac{ay^{b}z^{c}}{\lambda k}\right)^{\frac{a-1}{k-1}}$$
which then setting into the second equality to extract $y$ gives 
$$b\left(\frac{ay^{b}z^{c}}{\lambda k}\right)^{\frac{a-1}{k-1}}y^{b-1}z^{c}=\lambda ky^{k-1}$$
where I'm not sure how to even approach extracting $y$, and even if I will be able to seems to convoluted to be able to help me in the future..
Any better approaches to the question?
 A: Just played with it a little and got:
$$
\left(\begin{matrix}x^{a}y^{b}z^{c}\\
x^{a}y^{b}z^{c}\\
x^{a}y^{b}z^{c}
\end{matrix}\right)=\lambda k\left(\begin{matrix}x^{k}/a\\
y^{k}/b\\
z^{k}/c
\end{matrix}\right)
$$
Therefore $x^k/a=y^k/b=z^k/c$. By putting it in the constraint, we get
$$
1=x^k+y^k+z^k=(1+b/a+c/a)x^k \\
x=\sqrt[k]{\frac{a}{a+b+c}}
$$
and similarly $y=\sqrt[k]{\frac{b}{a+b+c}}$ and $z=\sqrt[k]{\frac{c}{a+b+c}}$.
A: $$ax^{a-1}y^bz^c=\lambda kx^{k-1}$$
$$ax^ay^bz^c=af(x,y,z)=\lambda k x^k$$
Similarly
$$bf(x,y,z)=\lambda ky^k$$
And one with $z$
Add all three together
$$(a+b+c)f(x,y,z)=\lambda k $$
$$f(x,y,z)=\frac {\lambda k}{a+b+c}$$
However after this part, I think it is quite impossible to find exactly how $\lambda$ depends on $a,b,c,k$. I have tried using wolfram alpha to check the results for different values of constants but I don't see any patterns. Most of the cases, using the above formula to trace back the value of $\lambda$, it is usually in a very messy surd expression. Maybe if given specific cases for $a,b,c,k$, one can find the value.
A: For those interested I have asked for help in a different place and was able to reach a solution. Both @Olorin and @Ido have found the same solution. The hint that allowed me to solve this was to try and maximize $\ln f$ instead of $f$. The complete solution is:

Let $g\left(x,y,z\right)=x^{k}+y^{k}+z^{k}-1$. We will search for the maximum of $\ln f$ constrained by $g\left(x,y,z\right)=0$. That is we are interested in $x,y,z$ where for some $\lambda$, 
  $$\nabla\left(\ln f\right)=\lambda\nabla g$$
As $$\ln f\left(x,y,z\right)=a\ln x+b\ln y+c\ln z$$
   we can calculate the derivative: $$\nabla\left(\ln f\right)=\begin{pmatrix}a/x\\
b/y\\
c/z
\end{pmatrix}$$
   and $$\nabla g=\begin{pmatrix}kx^{k-1}\\
ky^{k-1}\\
kz^{k-1}
\end{pmatrix}$$
hence we want to find $x,y,z$ and $\lambda$ such that
  $$\begin{pmatrix}a/x\\
b/y\\
c/z
\end{pmatrix}=\lambda\begin{pmatrix}kx^{k-1}\\
ky^{k-1}\\
kz^{k-1}
\end{pmatrix}$$
which gives us the equations
  $$\frac{a}{x} = \lambda kx^{k-1}
\frac{b}{y} = \lambda ky^{k-1}
\frac{c}{z} = \lambda kz^{k-1}$$
together with $$x^{k}+y^{k}+z^{k}=1$$
Solving the first set of equations for $x,y,z$ we see that
  $$x = \left(\frac{a}{\lambda k}\right)^{1/k}
y = \left(\frac{b}{\lambda k}\right)^{1/k}
z = \left(\frac{c}{\lambda k}\right)^{1/k}$$
  hence $$\frac{a}{\lambda k}+\frac{b}{\lambda k}+\frac{c}{\lambda k}=1$$
  which shows us that $$\lambda=\frac{a+b+c}{k}$$
   and that a critical point is at $$x=\left(\frac{a}{a+b+c}\right)^{1/k},\ y=\left(\frac{b}{a+b+c}\right)^{1/k}\text{ and }z=\left(\frac{c}{a+b+c}\right)^{1/k}$$
   where as $\ln$ is a monotone function, this must also be a critical point of $f$ on the set. Apart from this point all extrema must appear on the boundary, where one of $x,y,z$ is equal $0$, and thus $f\left(x,y,z\right)=0$ , where the set being compact shows that both a maximum and a minimum exist, and specifically this point is indeed a maximum, and the constrained maximum is $$f\left(\left(\frac{a}{a+b+c}\right)^{1/k},\left(\frac{b}{a+b+c}\right)^{1/k},\left(\frac{c}{a+b+c}\right)^{1/k}\right)=\left(\frac{a^{a}b^{b}c^{c}}{\left(a+b+c\right)^{a+b+c}}\right)^{1/k}$$

