Inequality for harmonic means

Question

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

Answer 1

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) $

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

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

Answer 2

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 $$\sum_{k=1}^n{k\over\sum_{j=1}^k 1/a_j}\leq c\sum_{k=1}^n a_k.$$

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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(n+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 $p$ $$\sum_{n=1}^\infty \left({n\over \sum_{j=1}^n1/a_j}\right)^p \leq\left({p+1\over p}\right)\sum_{n=1}^\infty a_n^p.$$

Answer 3

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 $2$.

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Sharpness

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.

Answer 4

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.