The maximum value? $x^5y + y^5z + z^5x$
Let $x$, $y$, and $z$ be non-negative reals such that $x+y+z=1$
The maximum value of the above expression can be represented as $a^b/c^d$, where a and c are not perfect powers, and $a, b, c, d$ are positive integers. Find the value of $a+b+c+d$.
I want to know how to solve such questions. What are useful methods? What should I learn about to solve these "maximum and minimum values"?
Note: I'm a self-learner. What happened is that I saw this question on a site and many other such questions about "maximum and minimum values". My main point is to know what should I learn about to solve it (Resources would be helpful).
 A: I don't know if the following answer is the better one, but it shows that $x = y = z = 1/3$ is not the optimal solution.
Let $f(x,y,z) = x^5y +y^5z+z^5x$ be the objective function. The problem is equivalent to minimize the Lagrangian function
$$ L =  -f(x,y,z) -\mu_1x-\mu_2y- \mu_3z + \lambda(x+y+z-1),$$ where $\mu_1, \mu_2, \mu_3$  and $\lambda$ are Lagrangian multipliers. The stationary points satisfy the polynomial system
$$ \lambda = \mu_1 + 5x^4y +z^5$$
$$ \lambda = \mu_2 + 5y^4z +x^5 $$
$$ \lambda = \mu_3 + 5z^4x + y^5 $$
$$ x + y + z =1$$
$$ \mu_1x = 0 $$ 
$$ \mu_2y = 0 $$ 
$$ \mu_3z = 0 .$$ 
$$ \mu_1 \geq 0,\mu_2 \geq 0,\mu_3 \geq 0$$
Now, we evaluate all possible cases for the multipliers $\mu_1, \mu_2$, and $\mu_3.$ 
Case (1): $\mu_1, \mu_2$, and $\mu_3$ are positives
In that case the system has no solution (the last three equations lead to $x=y=z=0$, not satisfying the equality constraint).
Case (2): at most a multiplier is zero
There is no solution such as all multipliers are nonnegative. Indeed, assuming $\mu_1 = 0$, we have $(x,y,z) = (1,0,0) \implies \lambda =0.$ Hence, $\mu_2 = -1$. Similarly, $\mu_2 = 0 \implies \mu_3 = -1$ or $\mu_3 = 0 \implies \mu_1 = -1$. 
Case (3): at most two multipliers are zero
The set of solutions are $$(x,y,z) = \{(5/6,1/6,0),(1/6,0, 5/6),(0,5/6,1/6)\},$$
for $\lambda \approx 0.4019$ in all solutions, and
$$(\mu_1,\mu_2,\mu_3) = \{(0,0, 0.4017),(0,0.4017, 0),(0.4017,0,0)\}.$$ The objective is $f(x,y,z) \approx 0.067.$
Case (4): all multipliers are zero
In this case, we need to solve the following system
$$ 5x^4y + z^5 = 5y^4z + x^5$$
$$ 5z^4x +y^5 = 5y^4z + x^5 $$
$$x +y +z =1, $$ which it seems to be not so simple. An idea is to compute a Grobner basis in order to simplify the system to tackle univariate polynomials. See the basis computed with WolframAlpha here.
Nevertheless,I've computed the solutions of the system with the command solve of Matlab. It gives a total of $25$ solutions, but only four ones are real and nonnegative. The solutions given by the solver are
$$(x,y,z) = \{(1/3,1/3,1/3), (0.3631,0.5426,0.0943), (0.5426,0.0943, 0.3631), (0.0943, 0.3631, 0.5426)\}.$$
For the first solution, the objective is $f(x,y,z) \approx 0.0041$. The other three solutions give the value $f(x,y,z) \approx 0.0079$.
Thus, the optimal solution lies in the case (3).
