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let $a_{i}>0,i=1,2,\cdots,n,n\ge 3$,show that $$\dfrac{n^2-1}{6}\min_{1\le i<j\le n}\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^2\le\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}-\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\le\dfrac{n-1}{n}\max_{1\le i<j\le n}\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^2$$

let $$A_{n}=\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n},G_{n}=\sqrt[n]{a_{1}a_{2}\cdots a_{n}}$$ then we have prove

$$\dfrac{n^2-1}{6}\min_{1\le i<j\le n}\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^2\le A_{n}-G_{n}\le\dfrac{n-1}{n}\max_{1\le i<j\le n}\left(\sqrt{a_{i}}-\sqrt{a_{j}}\right)^2$$

I found RHS if and only if $a_{k}=0,a_{1}=a_{2}=\cdots=a_{k-1}=a_{k+1}=\cdots=a_{n}$ .

and LHS if and only if $a_{k+1}-a_{k}=d$ ,where $d$ is constant,mean that $\{a_{n}\}$ is Arithmetic progression

I found sometimes and find this http://journals.math.tku.edu.tw/index.php/TKJM/article/viewFile/747/626

and http://arxiv.org/pdf/1203.4454.pdf

and http://files.ele-math.com/articles/jmi-03-21.pdf

and http://projecteuclid.org/download/pdf_1/euclid.rmjm/1181071887

and this paper prove this follow interesting result:

$$\dfrac{1}{2bn^2}\sum_{j<k}(x_{j}-x_{k})^2\le\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}-\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\le\dfrac{1}{2an^2}\sum_{j<k}(x_{j}-x_{k})^2$$

http://www.ams.org/journals/proc/1978-071-01/S0002-9939-1978-0476971-2/S0002-9939-1978-0476971-2.pdf

Martin Sleziak
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math110
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2 Answers2

5

We proceed by mixing variables. Without loss of generality, let $\max a_i=b$ and $\min a_i=a.$ Observe, that fixing the sum $S=\sum_{i=1}^n a_i$ we can choose two different $a_i\ge a_j$ and change them to $b\ge a_i'=a_i+\varepsilon$ and $a_j'=a_j-\varepsilon\ge a.$ The the left hand side of our inequality will increase and the right hand side remains the same. Applying this operation, we end up with $k$ numbers equal to $a,$ $n-k-1$ number equal to $b$ and $a\le a_{i_0}=c\le b.$ So we are left to maximize the function $$f(c)=\frac{ka+(n-k-1)b+c}{n}-\sqrt[n]{a^kb^{n-k-1}c.}$$ Since $f''(c)>0,$ the function $f$ is concave down and attains its maximum on the interval $[a,b]$ at one of the end points. Without loss of generality $c=b.$ Then, $$f(b)=\frac{ka+(n-k)b}{n}-\sqrt[n]{a^kb^{n-k}}\le \frac{n-1}{n}(a-2\sqrt{ab}+b)$$ which is equivalent to $$\frac{(n-1-k)a+(k-1)b+n\sqrt[n]{a^kb^{n-k}}}{2(n-1)}\ge \sqrt{ab}.$$ The latter immediately follows from Cauchy applied to $2n-2$ numbers. In exactly the same way one can prove the LHS of our inequality.

leshik
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3

By now, this is just a partial answer, proving the lower bound.

By following Aldaz, we have that it is a good idea to consider the variance of the set $X=\{\sqrt{a_1},\ldots,\sqrt{a_n}\}$. It happens that: $$ 0 \leq \operatorname{Var}(X)=\mathbb{E}[(X-\mathbb{E}[X])^2]=\mathbb{E}[X^2]-\mathbb{E}[X]^2\leq(AM-GM)(a_1,\ldots,a_n)$$ since $\mathbb{E}[X]^2$ is the order-$\frac{1}{2}$ mean, that is greater than the geometric mean.

Assume now that $a_1\leq a_2\leq\ldots\leq a_n$ and that $$\min_{2\leq i\leq n}\left(\sqrt{a_i}-\sqrt{a_{i-1}}\right)=L.$$ The lower bound $\operatorname{Var}(X)\geq\frac{n^2-1}{6}\,L$ now follows by supposing that $\mathbb{E}[X]$ lies between $\sqrt{a_k}$ and $\sqrt{a_{k+1}}$, by lower-bounding $\sqrt{a_{n+m}}-\sqrt{a_{n}}$ with $mL$ and minimizing on $k$.

Anyway, Aldaz' Theorem 2.2, point $(2)$, in http://arxiv.org/pdf/1203.4454.pdf, states that:

$$\frac{n}{n-1}\operatorname{Var}(\sqrt{a_1},\ldots,\sqrt{a_n})\leq (AM-GM)(a_1,\ldots,a_n)\leq n\operatorname{Var}(\sqrt{a_1},\ldots,\sqrt{a_n})$$

that is way stronger than the given inequality. The hardest part is the upper bound, achieved through induction on $n$ and Lagrange multipliers.

Jack D'Aurizio
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