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For symmetric functions, people ask Do symmetric problems have symmetric solutions?, e.g., [3] and [4]. The answer is no in general. However, solutions of symmetric problems often exhibit certain symmetry. In [1], for a class of symmetric polynomials, the global minimum under some conditions is attained at some points with $|\{x_1, x_2, \cdots, x_n\} |\le 2$, i.e., at most two distinct components. In [2], for a linear combination of elementary symmetric polynomials, under some conditions, each of the local extrema ($n$-dimensional vector) has at most $k$ distinct components.

First, let us see an example. Under the conditions $x, y, z > 0$ and $xyz = 1$, the global minimum of $$g(x,y,z) = \frac{\sin \frac{\pi x}{2} }{x} + \frac{\sin \frac{\pi y}{2}}{y} + \frac{\sin \frac{\pi z}{2}}{z}$$ is attained at some points $(x_0, y_0, z_0)$ with exactly two of $x_0, y_0, z_0$ being equal, for example $(x_0, x_0, \frac{1}{x_0^2})$ where $x_0 \approx 2.852$ and the minimum $g_{\min} \approx 0.878$. Also $g(1,1,1) = 3 > g_{\min}$ and $\lim_{\min(x,y,z)\to 0^{+}} g(x,y,z) > 0.884 > g_{\min}$.

In the problem above, the components of the global minimizer is not distinct. Now, suppose $f: (0, \infty) \rightarrow \mathbb{R}$ is a differentiable function. Let $F(x, y, z) = f(x) + f(y) + f(z)$. I want to find some examples of $f$ such that under the conditions $x, y, z>0$ and $xyz=1$, the global minimum of $F(x,y,z)$ is attained at some point $(x_0, y_0, z_0)$ with none of $x_0, y_0, z_0$ being equal, if not impossible.


Reference:

[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007. Online: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf

[2] Alexander Kovacec, et. al., “A note on extrema of linear combinations of elementary symmetric functions”, Linear and Multilinear Algebra, Volume 60, 2012 - Issue 2.

[3] R. F. Rinehart, "On Extrema of Functions which Satisfy Certain Symmetry Conditions", The American Mathematical Monthly, Vol. 47, No. 3 (Mar., 1940), pp. 145-152.

[4] William C. Waterhouse, “Do Symmetric Problems Have Symmetric Solutions?”, The American Mathematical Monthly, vol. 90, 1983, pp. 378-387.

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  • $\begingroup$ Isn't your recent edit an answer to your question? (Also, the no-components-$0$ seems to be a red herring, since, given any solution, you can shift all components by the same amount to make all components non-$0$ while preserving symmetry of the objective.) $\endgroup$
    – LSpice
    Commented Dec 24, 2019 at 1:59
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    $\begingroup$ @LSpice Thanks for comments. For cyclic symmetry rather than symmetry, we got an answer. Also, you are right about no-components-0, although the point is that all components are distinct. Other constraints (as long as they are symmetric) are allowed. $\endgroup$
    – River Li
    Commented Dec 24, 2019 at 2:53
  • $\begingroup$ Nice result sir $\endgroup$
    – user164469
    Commented Feb 20, 2021 at 11:58

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