For a fixed positive integer $n,$ compute the minimum value of the sum $$ x_{1}+\frac{x_{2}^{2}}{2}+\frac{x_{3}^{3}}{3}+\ldots+\frac{x_{n}^{n}}{n} $$ where $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are positive real numbers such that $$ \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\ldots+\frac{1}{x_{n}}=n $$

I've been trying this problem for days but couldn't solve it.

The author of the book from this question is taken expects the reader to solve it via R.M.S.-A.M.-G.M.-H.M. inequalities and Weighted A.M.-G.M. inequalities.

One is free to answer it through advanced inequalities (Cauchy - Schwarz, etc.) but solving it through inequalities of means is more preferred.

I have done some hopeless tries which are not worthy of being included here.

What I have thought is that the inequalities of means preserve the degree in the RHS and the LHS of an inequality so maybe using only Means is not sufficient, maybe we can use the square root to find a minimum value.

Any help would be appreciated.,


2 Answers 2


Firstly, from AM-GM:

$$x_k+\frac{1}{x_k}\geq 2\Rightarrow x_k\geq 2-\frac{1}{x_k}$$

Using AM-GM one more time:

$$x_k^k+(k-1)=x_k^k+\underbrace{1+1+\ldots+1}_{k-1\ \text{times}}\geq k\sqrt[k]{x_k^k}=kx_k\geq k\left(2-\frac{1}{x_k}\right)$$


$$\frac{x_k^k}{k}\geq 2-\frac{k-1}{k}-\frac{1}{x_k}=1+\frac{1}{k}-\frac{1}{x_k}$$

Summing, from $k=1$ to $n$:

$$\sum_{k=1}^n \frac{x_k^k}{k}\geq n+\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n \frac{1}{x_k}=\sum_{k=1}^n\frac{1}{k}$$

Equality occurs when $x_k=1$.

  • $\begingroup$ Is this the minimum value the sum can take? $\endgroup$
    – Crocogator
    Mar 10, 2020 at 7:44
  • $\begingroup$ Yes, since $n$ is given, the minimum value the sum can reach is $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}$, for which we don't have a closed form, but it's known as the harmonic sequence. This value is reached when $x_1=x_2=...=x_n=1$. $\endgroup$
    – LHF
    Mar 10, 2020 at 7:47
  • 1
    $\begingroup$ +1 ingenious... $\endgroup$ Mar 10, 2020 at 7:47
  • 1
    $\begingroup$ +1 Ok , I get it, thanks (•‿•) $\endgroup$
    – Crocogator
    Mar 10, 2020 at 7:48

In Application of weighted AM-GM it is shown that for non-negative real numbers $x_1, \ldots, x_n$ $$ \tag{*} x_1+\frac{x_2^2}{2}+\frac{x_3^3}{3}+\dots+\frac{x_n^n}{n}\ge H_n \cdot \left(x_1x_2x_3...x_n \right)^{1/H_n} $$ with $H_n = 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$, the $n^\text{th}$ harmonic number.

In our case $$ 1 = \frac{n}{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\ldots+\frac{1}{x_{n}}} \le \sqrt[n]{x_1 x_2 \cdots x_n } $$ by the inequality between harmonic and geometric mean. It follows that $x_1 \cdots x_n \ge 1$, and substituting that in $(*)$ gives $$ x_1+\frac{x_2^2}{2}+\frac{x_3^3}{3}+\dots+\frac{x_n^n}{n}\ge1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n} \, . $$ Equality holds for $x_1 = \ldots = x_n = 1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .