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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.

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    $\begingroup$ Simple Lagrange multiplier yields $x_k=C^{\frac 1{k+1}}$ where $C$ is determined from the constraint. $\endgroup$ – A.S. Jan 16 '16 at 4:24
  • $\begingroup$ Is there a way without Lagrange multipliers? $\endgroup$ – Puzzled417 Jan 16 '16 at 4:27
  • $\begingroup$ Lagrange multipliers are usually the way to go when optimizing with an equality constraint. $\endgroup$ – Ben Longo Jan 16 '16 at 4:35
  • $\begingroup$ Minimum is achieved at $x_i=1$ ($C=1$). $\endgroup$ – A.S. Jan 16 '16 at 4:36
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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 $x_1=\cdots=x_n=1$.

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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.

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