Let $a_1,a_2,a_3, \cdots ,a_n$ be positive real numbers. For the integers $n\ge 2$, prove that\[ \left (\frac{\sum_{j=1}^{n} \left (\prod_{k=1}^{j}a_k \right )^{\frac{1}{j}}}{\sum_{j=1}^{n}a_j} \right )^{\frac{1}{n}}+\frac{\left (\prod_{i=1}^{n}a_i \right )^{\frac{1}{n}}}{\sum_{j=1}^{n} \left (\prod_{k=1}^{j}a_k \right )^{\frac{1}{j}}}\le \frac{n+1}{n}\]
Problem
Source: China Hangzhou
Tags: inequalities, inequalities proposed, China TST
19.03.2015 21:40
Though I haven't actually attempted either problem, I would like to point out the glaring similarity to SL 2004 A7. Does anyone know if the solution there can be easily adapted to solve the TST problem?
20.03.2015 00:46
It appears to be similar to the Mixed Arithmetic-Geometric Inequality by Kiran Kedlaya. If we can combine this and the above ISL 2004 we arrive at this inequality.
23.03.2015 05:04
After failing this for a VERY LONG TIME, I quit and googled the article about mixed AM-GM. A VERY amusing note: Kiran Kedlaya says that the result shown in the article is a conjecture of F. Holland, who in fact proposed 2004 A7. Hmmm.
25.03.2015 21:00
Could someone please write up a solution and post it? Thank you very much !
26.03.2015 07:19
My solution: Let $ \mathcal{G}_j= ( \prod_{i=1}^{j} a_i )^{\frac{1}{j}} $ ( $ \forall j=1,2, \dots , n $ ), so $ \sum_{j=1}^{n} a_j=\mathcal{G}_1+\frac{{\mathcal{G}_2}^2}{\mathcal{G}_1}+ \dots + \frac{{\mathcal{G}_n}^n}{{\mathcal{G}_{n-1}}^{n-1}} $ . Let $ \mathbb{G}=\mathcal{G}_1+\mathcal{G}_2+ \dots + \mathcal{G}_n $ and $ \mathbb{G}'=\mathcal{G}_1+\mathcal{G}_2+ \dots + \mathcal{G}_{n-1} $ . Let $ \mathbb{G}_t=\mathcal{G}_1+\mathcal{G}_2+ \dots+\mathcal{G}_{t-1}+\mathcal{G}_t+\mathcal{G}_t+\mathcal{G}_{t+1} +\dots+ \mathcal{G}_{n-1} $ ( $ \forall t=1,2, \dots , n-1 $ ) . From AM-GM inequality we get $ \left ( \frac{n \cdot \mathbb{G}'}{n-1} \right )^{n-1} \ge \mathbb{G}_1 \cdot \mathbb{G}_2 \cdot \dots \cdot \mathbb{G}_{n-1} $ . ... $ (1) $ From Holder inequality we get $ \mathbb{G}_1 \cdot \mathbb{G}_2 \cdot \dots \cdot \mathbb{G}_{n-1} \cdot ( \mathcal{G}_1+\frac{{\mathcal{G}_2}^2}{\mathcal{G}_1}+ \dots + \frac{{\mathcal{G}_n}^n}{{\mathcal{G}_{n-1}}^{n-1}} ) \ge {\mathbb{G} }^n $ . ... $ (2) $ From $ (1) , (2) $ we get $ \left ( \frac{n \cdot \mathbb{G}'}{n-1} \right )^{\frac{n-1}{n}} \left ( \mathcal{G}_1+\frac{{\mathcal{G}_2}^2}{\mathcal{G}_1}+ \dots + \frac{{\mathcal{G}_n}^n}{{\mathcal{G}_{n-1}}^{n-1}} \right )^{\frac{1}{n}} \ge \mathbb{G} $ $ \Longrightarrow \left ( \frac{ n \cdot \mathbb{G}'}{(n-1) \cdot \mathbb{G}} \right )^{\frac{n-1}{n}} \ge \left ( \frac{\mathbb{G}}{\mathcal{G}_1+\frac{{\mathcal{G}_2}^2}{\mathcal{G}_1}+ \dots + \frac{{\mathcal{G}_n}^n}{{\mathcal{G}_{n-1}}^{n-1}}} \right )^{\frac{1}{n}} $ . ... $ (3) $ Notice that $ (n-1) \cdot \frac{\mathbb{G}'}{(n-1) \cdot \mathbb{G}}+\frac{1}{n} \ge n \cdot \left ( \frac{{\mathbb{G}'}^{n-1}}{n \cdot {\mathbb{G}}^{n-1} \cdot (n-1)^{n-1}} \right )^{\frac{1}{n}}=\left ( \frac{ n \cdot \mathbb{G}'}{(n-1) \cdot \mathbb{G}} \right )^{\frac{n-1}{n}} $ . ... $ (4) $ From $ (3), (4) $ we get $ \text{L.H.S} \leq \frac{\mathbb{G}'}{\mathbb{G}} +\frac{1}{n}+\frac{\mathbb{G}-\mathbb{G}'}{\mathbb{G}}=\frac{n+1}{n}=\text{R.H.S} $ . Q.E.D
29.03.2015 17:27
Congrats on an amazing solution Telvcohl!
29.03.2015 18:12
Well done , kid !
19.03.2020 04:44
Is this solution equivalent to TelvCohl's? Again set $G_j = \left( \prod_{i=1}^{j} a_i \right)^{\frac{1}{j}} ( \forall j=1,2, \dots , n )$, so that: $$a_j = \frac{G_j^j}{G_{j-1}^{j-1}}$$Then upon rearranging it suffices to show: $$\left (\frac{\sum_{j=1}^{n} \left (\prod_{k=1}^{j}a_k \right )^{\frac{1}{j}}}{\sum_{j=1}^{n}a_j} \right )^{\frac{1}{n}}+\frac{\left (\prod_{i=1}^{n}a_i \right )^{\frac{1}{n}}}{\sum_{j=1}^{n} \left (\prod_{k=1}^{j}a_k \right )^{\frac{1}{j}}}\le \frac{n+1}{n}$$$$\iff \left (\frac{\sum_{j=1}^{n} G_j }{\sum_{j=1}^{n}a_j} \right )^{\frac{1}{n}} \le \frac{n+1}{n} - \frac{G_n}{\sum_{j=1}^{n} G_j}$$$$\iff n^n \left (\sum_{j=1}^{n} G_j \right )^{n+1} \le \left(\sum_{j=1}^{n-1} (n+1)G_j + G_n \right)^{n}\sum_{j=1}^{n}a_j$$$$\iff n^n \left (\sum_{j=1}^{n} G_j \right )^{n+1} \le \left(\sum_{j=1}^{n-1} (n+1)G_j + G_n \right)^{n} \left(\sum_{j=1}^{n} \frac{G_j^j}{G_{j-1}^{j-1}} \right)$$From AM-GM: $$\left(\sum_{j=1}^{n-1} (n+1)G_j + G_n \right)^n = \left(\sum_{j=1}^{n} \left[(j-1)G_{j-1} + (n-j+1)G_{j}\right] \right)^n \geq n^n \left( \sum_{j=1}^{n} G_{j-1}^{\frac{j-1}{n}}G_{j}^{\frac{n-j+1}{n}} \right)^n$$From Holder's: $$\left( \sum_{j=1}^{n} G_{j-1}^{\frac{j-1}{n}}G_{j}^{\frac{n-j+1}{n}} \right)^n \left(\sum_{j=1}^{n} G_{j-1}^{1-j}G_j^j \right) \geq \left( \sum_{j=1}^{n} G_{j} \right)^{n+1}$$So by multiplying these two inequalities we get the desired result. $\blacksquare$
04.01.2021 17:02
$G_j= ( \prod_{i=1}^{j} a_i )^{\frac{1}{j}} $and$a_j = \frac{G_j^j}{G_{j-1}^{j-1}}$ Asume that $\sum_{j}^n G_j=1$ We only need to prove $$\left (\frac{\sum_{j=1}^{n} G_j }{\sum_{j=1}^{n}\frac{G_j^j}{G_{j-1}^{j-1}}} \right )^{\frac{1}{n}} \le \frac{n+1}{n} - \frac{G_n}{\sum_{j=1}^{n} G_j}$$$$\left (\sum_{j=1}^{n} \frac{ G_j^j }{G_{j-1}^{j-1}} \right )^{-\frac{1}{n}} =\left (\sum_{j=1}^{n} G_j\frac{ G_j^{j-1} }{G_{j-1}^{j-1}} \right )^{-\frac{1}{n}} \le \sum_{j=1}^{n} G_j\left (\frac{ G_j^{j-1} }{G_{j-1}^{j-1}} \right )^{-\frac{1}{n}}$$(Weighed Jensen) $$\sum_{j=1}^{n} G_{j-1}^{1-\frac{j-1}{n}}G_j^{\frac{j}{n}} \le \sum_{j=1}^{n} (1-\frac{j-1}{n})G_{j-1}+\frac{j}{n}G_j$$(Weighed AM-GM) $=\frac{n+1}{n} - \frac{G_n}{\sum_{j=1}^{n} G_j}$ Hence,we have done.
17.05.2023 06:49
For $\forall 1\leq j\leq n$, define $G_j=\sqrt[j]{\prod\limits_{k=1}^ja_k}$, let $\sum_{j=1}^nG_j=1$. Then $$\Leftrightarrow\left(\sum\limits_{j=1}^n\dfrac{G_j^j}{G_{j-1}^{j-1}}\right)^{-\frac 1n}+G_n\leqslant\frac{n+1}n.$$Using Jensen inequality, $$\left(\sum\limits_{j=1}^n\dfrac{G_j^j}{G_{j-1}^{j-1}}\right)^{-\frac 1n}=\left(\sum\limits_{j=1}^nG_j\cdot\dfrac{G_j^{j-1}}{G_{j-1}^{j-1}}\right)^{-\frac 1n}\leqslant\sum_{j=1}^nG_j\left(\dfrac{G_j^{j-1}}{G_{j-1}^{j-1}}\right)^{-\frac 1{n}}=\sum_{j=1}^nG_{j-1}^{\frac{j-1}n}G_j^{1-\frac{j-1}n}.$$Using AM-GM inequality, $$\sum_{j=1}^nG_{j-1}^{\frac{j-1}n}G_j^{1-\frac{j-1}n}\leqslant\sum_{j=1}^n\left[\frac{j-1}nG_{j-1}+\left(1-\frac{j-1}n\right)G_j\right]=\frac{n+1}n\sum_{j=1}^nG_j-G_n=\frac{n+1}n-G_n.$$Therefore $$\left(\sum\limits_{j=1}^n\dfrac{G_j^j}{G_{j-1}^{j-1}}\right)^{-\frac 1n}+G_n\leqslant\frac{n+1}n.\blacksquare$$
07.09.2023 12:36
With the same notation set $G_j=\sqrt[j]{\prod_{k=1}^{j}a_j}$, we need to prove $$\left (\frac{\sum_{j=1}^{n} G_j }{\sum_{j=1}^{n}\frac{G_j^j}{G_{j-1}^{j-1}}} \right )^{\frac{1}{n}} \le \frac{n+1}{n} - \frac{G_n}{\sum_{j=1}^{n} G_j}=\frac{1}{n}+(n-1)\times \frac{\sum_{j=1}^{n-1}G_j}{(n-1)\sum_{j=1}^{n}G_j}$$Now by AM-GM we have $$\frac{1}{n}+(n-1)\times \frac{\sum_{j=1}^{n-1}G_j}{(n-1)\sum_{i=1}^{n}G_j}\ge \sqrt[n]{(\frac{n}{n-1})^{n-1}(\frac{\sum_{j=1}^{n-1}G_j}{(\sum_{j=1}^{n}G_j})^{n-1}}$$so we need to prove $$(\sum_{j=1}^{n}G_j)^n\le (\frac{n}{n-1}\sum_{j=1}^{n-1}G_j)^{n-1}(\sum_{j=1}^{n}\frac{G_j^j}{G_{j-1}^{j-1}})$$We want to use Holder-Carlson inequality so we need to rewrite $$\frac{n}{n-1}\sum_{j=1}^{n-1}G_j=\sum_{j=1}^{n}\frac{(j-1)G_{j-1}+(n-j)G_{j}}{n-1}$$Notice that $$(\frac{(j-1)G_{j-1}+(n-j)G_{j}}{n-1})^{n-1}\frac{G_j^j}{G_{j-1}^{j-1}}\ge G_{j-1}^{j-1}G_j^{n-j}\frac{G_j^j}{G_{j-1}^{j-1}}=G_j^n$$So by Holder-Carlson we are done! XDD:Just find my solution is equivalent to the above one...