Prove that for real numbers $a_1 ,a_2 ,...,a_n >0$ the following inequality holds $$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n }\leq 4\cdot \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right).$$

  • 2
    $\begingroup$ Just a comment .. you can replace the $4$ with $e$, there .. since $a_i > 0$ make the transformation $a_i \rightarrow \dfrac{1}{a_i}$, then use $HM \le GM$ inequality for the LHS, followed by Carleman's Inequality .. cheers :-) $\endgroup$ – r9m Jun 24 '14 at 7:58
  • $\begingroup$ @StevenTaschuk $e$ can be replaced by $2$ :-) $\endgroup$ – r9m Jun 24 '14 at 19:28
  • 4
    $\begingroup$ With $2$ instead of $4$, and $a_j$ replaced by $1/a_j$, this is American Mathematical Monthly problem 11145 published in the April 2005 issue, with the solution appearing in the October 2006 issue. The published solution is essentially the same as those below. They also show that the constant $2$ is sharp. $\endgroup$ – user940 Jun 25 '14 at 17:29
  • $\begingroup$ @ByronSchmuland: I have added a demonstration of the sharpness of $2$. Is is similar to the method used in AMM 11145? $\endgroup$ – robjohn Jun 25 '14 at 23:30
  • $\begingroup$ @robjohn I've added the AMM solution below. $\endgroup$ – user940 Jun 25 '14 at 23:50

The linked proof of Carleman's Inequality (in the comment) indicates the method of balancing coefficients in weighted mean inequalities. In the same spirit we can show, $$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n } < 2 \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right)$$ for positive $a_i$'s.

We choose a set of positive real numbers $x_1,x_2,\ldots,x_n$ such that, by Cauchy-Schwarz Inequality: $$(a_1+a_2+\cdots+a_k)\left(\dfrac{x_1^2}{a_1}+\dfrac{x_2^2}{a_2}++\cdots\dfrac{x_k^2}{a_k}\right) \ge (x_1+x_2+\cdots+x_k)^2$$

$$\implies \dfrac{k}{a_1+a_2+\cdots+a_k} \le \dfrac{k}{(x_1+x_2+\cdots+x_k)^2}\left(\dfrac{x_1^2}{a_1}+\dfrac{x_2^2}{a_2}++\cdots\dfrac{x_k^2}{a_k}\right)$$

for each $k=1,2,\cdots,n$.

Adding them up from $k=1$ to $n$, we get:

$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n } \le \dfrac{c_1}{a_1} + \frac{c_2}{a_2}+\ldots+\frac{c_n}{a_n}$$

Where, $c_k = \dfrac{kx_k^2}{(x_1+x_2+\cdots+x_k)^2} + \dfrac{(k+1)x_k^2}{(x_1+x_2+\cdots+x_{k+1})^2}+\ldots+\dfrac{nx_k^2}{(x_1+x_2+\cdots+x_n)^2}$ for each $k=1,2,\ldots,n$.

It remains to choose a set of $\{x_k\}_{k=1}^n$ such that $\max\limits_{k \in \{1,2,\cdots,n\}}\{c_k\}$ is minimized.

For example if we plug in $x_k = k$, we have,

$c_k = k^2\left(\sum\limits_{j=k}^n \dfrac{j}{(1+2+\cdots+j)^2}\right) = 4k^2\left(\sum\limits_{j=k}^n \dfrac{1}{j(j+1)^2}\right) $

$$\le 2k^2\left(\sum\limits_{j=k}^n \dfrac{2j+1}{j^2(j+1)^2}\right) = 2k^2\left(\sum\limits_{j=k}^n \dfrac{1}{j^2} - \sum\limits_{j=k}^n \dfrac{1}{(j+1)^2}\right)$$

$$ = 2k^2\left(\dfrac{1}{k^2} - \dfrac{1}{(n+1)^2}\right) < 2$$

  • 1
    $\begingroup$ Ack! I had my window open on this and the fact that your answer was posted never showed up until after I answered the captcha. Is my answer different enough? It seems simpler, at least. $\endgroup$ – robjohn Jun 25 '14 at 17:26
  • $\begingroup$ @robjohn Seems we started with the same thing in mind ^_^ ... I was just experimenting if the balancing coefficients in C-S could improve upon $2$ :-) $\endgroup$ – r9m Jun 25 '14 at 22:17
  • 1
    $\begingroup$ I have tried a little to show that $2$ is optimal. $\endgroup$ – robjohn Jun 25 '14 at 22:21

Upper Bound

By Cauchy-Schwarz, we have $$ \begin{align} \left(\sum_{j=1}^ka_j\right)\left(\sum_{j=1}^k\frac{j^2}{a_j}\right) &\ge\left(\sum_{j=1}^kj\right)^2\\[3pt] &=\frac{k^2(k+1)^2}{4}\tag{1} \end{align} $$ Thus, $$ \begin{align} \sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^ka_j} &\le\sum_{k=1}^n\frac4{k(k+1)^2}\sum_{j=1}^k\frac{j^2}{a_j}\\ &=\sum_{j=1}^n\frac{j^2}{a_j}\sum_{k=j}^n\frac4{k(k+1)^2}\\ &\le\sum_{j=1}^n\frac{j^2}{a_j}\sum_{k=j}^n2\left(\frac1{k^2}-\frac1{(k+1)^2}\right)\\ &\le\sum_{j=1}^n\frac{j^2}{a_j}\frac2{j^2}\\ &=2\sum_{j=1}^n\frac1{a_j}\tag{2} \end{align} $$ Thus, the ratio is at most $2$.


Set $a_k=k^\beta$ for $0\lt\beta\lt1$. First $$ \begin{align} \sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^ka_j} &=\sum_{k=1}^n\frac{k}{\frac1{1+\beta}k^{\beta+1}+O(k^\beta)}\\ &=(1+\beta)\sum_{k=1}^n\frac1{k^\beta+O(k^{\beta-1})}\\ &=\frac{1+\beta}{1-\beta}n^{1-\beta}+O(1)\\[3pt] &\sim\frac{1+\beta}{1-\beta}n^{1-\beta}\tag{3} \end{align} $$ Next $$ \begin{align} \sum_{k=1}^n\frac1{a_k} &=\sum_{k=1}^n\frac1{k^\beta}\\ &=\frac1{1-\beta}n^{1-\beta}+O(1)\\[3pt] &\sim\frac1{1-\beta}n^{1-\beta}\tag{4} \end{align} $$ Thus, as $\beta\to1^-$, the ratio tends to $2$ for large $n$. Therefore, $2$ is sharp.


AMM problem 11145 (April 2005)

Proposed by Joel Zinn, Texas A&M University, College Station, TX.

Find the least $c$ such that if $n\geq 1$ and $a_1,\dots,a_n>0$, then $$\sum_{k=1}^n{k\over\sum_{j=1}^k 1/a_j}\leq c\sum_{k=1}^n a_k.$$

A Sum Inequality (October 2006)

Solution by Grahame Bennett, Indiana University, Bloomington, IN.

Let $S_n$ denote the sum on the left hand side of the proposed inequality. The Cauchy-Schwarz inequality gives $(\sum_1^k j)^2\leq \sum^k_1j^2a_j\, \sum^k_1 1/a_j$, or equivalently, $${k\over\sum_{j=1}^k 1/a_j}\leq{4\over k(k+1)^2}\sum_{j=1}^k j^2a_j.$$ Summing over $k$ yields \begin{eqnarray*} S_n&\leq&2 \sum_{j=1}^nj^2a_j\sum_{k=j}^n{2\over k(k+1)^2}\leq 2 \sum_{j=1}^nj^2a_j\sum_{k=j}^n{2k+1\over k^2(k+1)^2} \\ & = & 2 \sum_{j=1}^nj^2a_j\sum_{k=j}^n\left({1\over k^2}-{1\over(k+1)^2}\right) = 2 \sum_{j=1}^n j^2 a_j \left({1\over j^2}-{1\over(j+1)^2}\right) < 2\sum_{j=1}^n a_j, \end{eqnarray*} from which it follows that the stated inequality holds with $c=2$.

To see that no smaller value of $c$ is possible, we set $a_j=1/j$, in which case the left side is $\sum_{k=1}^n 2/(k+1)$ and the right side is $c\sum_{k=1}^n1/k$. Since the harmonic series diverges, we must have $c\geq 2$.

Editorial comment. The upper bound is a special case of a theorem in K. Knopp's article "Uber Reihen mit positiven Gliedern," J. London Math. Soc. 3 (1928), 205--211. It states that for positive $p$ $$\sum_{n=1}^\infty \left({n\over \sum_{j=1}^n1/a_j}\right)^p \leq\left({p+1\over p}\right)\sum_{n=1}^\infty a_n^p.$$


In this question: A version of Hardy's inequality involving reciprocals.

we prove the following:

$$ \frac{n^2}{2(a_1 + a_2 + \dots + a_n)} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$

which easily implies what you seek.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy