<s>[INST] Minimizing Sum of Reciprocals Find the minimum value, in terms of $k$ of $\frac{1}{x_1}+…+\frac{1}{x_n}$ if $x_1^2+x_2^2+…+x_n^2=n$ and $x_1+x_2+…+x_n=k$, where $\sqrt{n} < k \leq n$. 
I tried the am-hm, but how to relate with the sum of squares?
 [/INST] Checking the case $n = 2$
First condition:
$$
g(x,y) = x + y = k \iff y = - x + k
$$
Second condition:
$$
h(x,y) = x^2 + y^2 = 2 \iff y = \pm \sqrt{2 - x^2}
$$
This gives
\begin{align}
2 &= x^2 + (-x + k)^2=2 x^2 -2kx + k^2 \iff \\
1 &= x^2 - kx + k^2/2 = (x - k/2)^2 + k^2/4 \iff \\
x &= \frac{k\pm\sqrt{4-k^2}}{2} \quad \wedge \\
y &= \frac{-k\mp\sqrt{4-k^2}}{2} + k = \frac{k\mp\sqrt{4-k^2}}{2}
\end{align}
We got 2 points or 1 point ($k=2$) which satisfy the conditions, which agrees with an intersection of a line ($g=k$) and a circle ($h=2$). Their coordinates were expressed in terms of $k$. Inserting those points into $f$ we get:
\begin{align}
f(x,y) &= \frac{1}{x} + \frac{1}{y} \\
&= \frac{2}{k\pm\sqrt{4-k^2}} + \frac{2}{k\mp\sqrt{4-k^2}} \\
&= \frac{4k}{k^2 - (4 - k^2)} \\
&= \frac{2k}{k^2 - 2} \\
&= F(k)
\end{align}
This function is constant regarding $x$ and $y$, which includes the minimum.
(in progress)
The Lagrange function is
$$
L(x,\lambda, \mu) = f(x) + 
\lambda \left( g(x) - k \right) +
\mu \left( h(x) - n \right)
$$
Gradient components are
\begin{align}
\frac{\partial L}{\partial x_j} &= -\frac{1}{x_j^2} + \lambda + 2 \mu x_j
\quad (j\in\{1,\ldots,n\}) \\
\frac{\partial L}{\partial\lambda} &= g(x) - k \\
\frac{\partial L}{\partial\mu} &= h(x) - n
\end{align}
At critical points $x^*$ the above components vanish. So
$$
f(x^*) = \sum_i \frac{1}{x^*_i} 
= \sum_i \lambda x^*_i + 2 \mu (x^*_i)^2
= \lambda k + 2 \mu n
$$
The question is how to calculate $\lambda$ and $\mu$.
We have $n+2$ equations with $n+2$ unknowns, so there is a chance for a unique solution $(x, \lambda, \mu)$, but so far I found no way to extract it.
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