Minima of symmetric polynomials subject to two symmetric constraints The homogeneous symmetric polynomial of degree $k$ in $n$ variables is 
$$
f_k(x_1,x_2,\dots,x_n) = \sum_{i_1<i_2<\cdots<i_k}x_{i_1}x_{i_2}\cdots x_{i_k}.
$$
Consider the following constrained optimization problem: For $k\in\{2,\dots,n\}$,
$$
\text{minimize: }\,\,f_k(x_1,\dots,x_n)
\qquad
\text{ subject to: }\,\, \sum_{i=1}^n x_i = 1\,\, \text{ and }\,\,\sum_{i=1}^n \sqrt{x_i} = c
$$
for some constant $1\leq c\leq \sqrt{n}$. 
I want to characterize the form of the points $\mathbf{x}_{\text{opt}}$ that minimize this. 
If $1\leq c\leq \sqrt{k-1}$, the minimum is zero and is achieved by some point of the form $$
\mathbf{x}_{\text{opt}} =(x_1,\dots,x_{k-1},0,\dots,0)$$
with $n-k+1$ entries equal to zero. 
However if $\sqrt{k-1}\leq c\leq \sqrt{n}$, I don't know what to do. Numerical results show that the minimum will always be achieved by some point of the form
$$
\mathbf{x}_{\text{opt}} = (\underbrace{s,\dots,s}_{m\text{ times}}, \underbrace{t,\dots,t}_{n-m\text{ times}}\Bigr)
$$
for some $s,t\in[0,1]$ and $m\in\{1,\dots,n\}$, where the $m$, $s$, and $t$ will be different depending on $k$ and $c$. That is, the optimal vector will have at most two different elements $s$ and $t$. 
But how might I prove this??? Note that 
$$
\frac{d}{dx_i}f_{k}(x_1,\dots,x_n)= f_{k-1}(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n)
$$
so my attempt at Lagrange multipliers means that the optimal $\mathbf{x}$ must satisfy
$$
f_{k-1}(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n) = \lambda \frac{1}{\sqrt{x_i}} + \mu
$$
for some constants $\lambda$ and $\mu$ and for all $i$. But then I am stuck. Any help is appreciated.
 A: (At the moment, this is not a complete answer. It is only my attempt to prove something I suggested in the comments.)
Theorem 1. 
If $\vec{x}=(x_1,\dots,x_n)$ is a minimum, then at least one of $x_i$ is zero or  the entries of $\vec{x}$ take at most two distinct values.
The result of the Theorem will follow directly from the following lemma.
Lemma 2. If $x_4,\dots,x_n$ are held constant and we optimize only over $x_1$, $x_2$ and $x_3$, a critical point of $f_k(x_1,\dots,x_n)$ cannot be achieved if either one of $x_1$, $x_2$ or $x_3$ is zero or $x_1$, $x_2$ and $x_3$ are all different.
Proof.  Consider holding $x_4,\dots,x_n$ constant and varying over $x_1$, $x_2$, and $x_3$ subject to 
\begin{equation}\label{eq1}
x_1+x_2+x_3 = X 
\qquad\text{ and }\qquad
\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3} = C.\tag{1}
\end{equation}
For a fixed $x_1$, there is only one possible choice for $x_2$ and $x_3$ (up to a permutation of $x_2$ and $x_3$), and these values vary analytically as $x_1$ changes. Taking the derivatives of \eqref{eq1} with respect to $x_1$, we find
$$
1+\frac{dx_2}{dx_1} + \frac{dx_3}{dx_1} = 0  \qquad\text{ and }\qquad \frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\frac{dx_2}{dx_1} + \frac{1}{\sqrt{x_3}}\frac{dx_3}{dx_1} = 0. 
$$
Solving for $\frac{dx_2}{dx_1}$ and $\frac{dx_3}{dx_1}$ we find
$$
\frac{dx_2}{dx_1} = \frac{\sqrt{x_2}}{\sqrt{x_1}}\frac{\sqrt{x_3}-\sqrt{x_1}}{\sqrt{x_2}-\sqrt{x_3}}
\qquad\text{ and }\qquad
\frac{dx_3}{dx_1} = \frac{\sqrt{x_3}}{\sqrt{x_1}}\frac{\sqrt{x_2}-\sqrt{x_1}}{\sqrt{x_3}-\sqrt{x_2}}.
$$
Note that we can write
\begin{align}
f_k(x_1,\dots,x_n) =& x_1x_2x_3 f_{k-3}(x_4,\dots,x_n)\\
                    &\qquad + (x_1x_2 + x_1x_3 + x_2x_3) f_{k-2}(x_4,\dots,x_n) \\ 
                    &\qquad+ \underbrace{(x_1+x_2+x_3)}_{=X}f_{k-1}(x_4,\dots,x_n) + f_{k}(x_4,\dots,x_n) .\tag{2}\label{eq2}
\end{align}
The last line in \eqref{eq2} is constant. Taking the derivative of the entire expression in \eqref{eq2} with respect to $x_1$, we find
\begin{align}
\frac{d}{dx_1}f_k(\vec{x}) =
  &         \left(x_2x_3+x_1x_2\frac{dx_3}{dx_1} + x_1x_3\frac{dx_3}{dx_1} \right)f_{k-3}(x_4,\dots,x_n) + \left(x_2+x_3+(x_1+x_3)\frac{dx_2}{dx_1} + (x_1+x_2)\frac{dx_3}{dx_1} \right)f_{k-2}(x_4,\dots,x_n)\nonumber\\
  =&  \frac{\left(\sqrt{x_1}-\sqrt{x_3}\right)\left(\sqrt{x_1}-\sqrt{x_2}\right)}{\sqrt{x_1}}\bigl(\sqrt{x_1x_2x_3} f_{k-3}(x_4,\dots,x_n) - \left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\right)f_{k-2}(x_4,\dots,x_n)\bigr)\tag{3}\label{eq3}.
\end{align}
There is a critical point if and only if the derivative in \eqref{eq3} vanishes. If we suppose that $x_1$, $x_2$, and $x_3$ are all non-zero and distinct, then the derivative in \eqref{eq3} will vanish if and only if 
$$
\sqrt{x_1x_2x_3} f_{k-3}(x_4,\dots,x_n) - \bigl(\underbrace{\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}}_{=C}\bigr)f_{k-2}(x_4,\dots,x_n) = 0. \tag{4}\label{eq4}
$$
HOWEVER... This is where I get stuck. It still needs to be shown that \eqref{eq4} cannot vanish if $x_1$, $x_2$, $x_3$ are all nonzero and distinct. 
Furthermore, it can be shown that on the boundary of the allowable points $x_1,x_2,x_3$, either one of them is zero or two of them are equal.
Hence, if all of $x_1$, $x_2$, and $x_3$ are nonzero and distinct, then that point is NOT a minimum since it is neither a boundary point nor a critical point.
It follows that, if $\vec{x}=(x_1,\dots,x_n)$ is a minimum then at least one of $x_i$ is zero, or the entries of $\vec{x}$ take at most two distinct values.
