Find the minimum of $\sum_{k=1}^n \frac{x^k_k}{k}$ 
Let $n$ be a positive integer. Find the minimum of $\displaystyle \sum_{k=1}^n \dfrac{x^k_k}{k}$, where $x_1,x_2,\ldots,x_n$ are positive real numbers such that $\displaystyle \sum_{k=1}^n \dfrac{1}{x_k} = n$.

This question sort of reminds me of Cauchy-Schwarz but the fact that the denominators of the fractions in $\displaystyle \sum_{k=1}^n \dfrac{x^k_k}{k}$ are in increasing harmonic sequence, it seems hard to relate it. I can't really think of anything to do here.
 A: Lagrange multipliers are very straightforward for this problem: we get the equations  $$ x_j^{j-1}+\frac \lambda  {x_j^2}=0,\ \ \ \ j=1,\ldots n. $$ Thus $x_1^2=x_2^3=\cdots =x_n ^{n+1}$, and so 
$$
n=\sum_{k=1}^n\frac1 {x_k}=\sum_{k=1}^n \frac1 {x_1^{2/(k+1)}}.
$$
If $x_1>1$, all terms in the sum are less than one and the equality is impossible; and if $x_1 <1$, all terms in the sum are greater than one, so again no equality. It follows that
$x_1=\cdots=x_n=1$.
A: The constraint
$$
\sum_{k=1}^n\frac1{x_k}=n\tag{1}
$$
implies that any infinitesimal change $\delta x_k$ satisfies
$$
\sum_{k=1}^n\frac{\delta x_k}{x_k^2}=0\tag{2}
$$
To minimize
$$
\sum_{k=1}^n\frac{x_k^k}{k}\tag{3}
$$
we need to find $x_k$ so that any $\delta x_k$ that satisfies $(2)$, gives
$$
\sum_{k=1}^nx_k^{k-1}\,\delta x_k=0\tag{4}
$$
If $x_k^{k-1}$ is orthogonal to the same $\delta x_k$ that $\frac1{x_k^2}$ is, then there is a $\lambda$ so that
$$
x_k^{k-1}=\lambda\frac1{x_k^2}\tag{5}
$$
which means
$$
x_k=\lambda^{\frac1{k+1}}\tag{6}
$$
Combining $(1)$ and $(6)$ gives
$$
\sum_{k=1}^n\lambda^{-\frac1{k+1}}=n\tag{7}
$$
Equation $(7)$ has the solution $\lambda=1$. Thus, the extreme value is
$$
\begin{align}
\sum_{k=1}^n\frac{x_k^k}{k}
&=\sum_{k=1}^n\frac{\lambda^{\frac{k}{k+1}}}{k}\\[3pt]
&=H_n\tag{8}
\end{align}
$$
where $H_n$ is the $n^{\text{th}}$ Harmonic Number.
