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?

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.

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.

Could someone please write up a solution and post it? Thank you very much !

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

Congrats on an amazing solution Telvcohl!

Well done , kid !

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. $\blacksquare$

$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.

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 $$\left(\sum\limits_{j=1}^n\dfrac{G_j^j}{G_{j-1}^{j-1}}\right)^{-\frac 1n}+G_n\leqslant\frac{n+1}n.\blacksquare$$

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...