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Then the quality of the inequalities proposed that year wasn't very extraordinary if this is the most challenging. Write it in the form $$n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}+\sqrt[n]{(2-A_1/A_2)...(n-(n-1)A_{n-1}/A_n)}\leq n+1.$$ Use the AM-GM inequality to deduce that the second term of the LHS is at most $$\frac{1+(2-A_1/A_2)+...+(n-(n-1)A_{n-1}/A_n)}{n}.$$ Thus the inequality reduces to $$\frac{n+1}{2}+\frac{A_1/A_2+...+(n-1)A_{n-1}/A_n}{n} \geq n\cdot\sqrt[n^2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}.$$ Now, apply generalized AM-GM inequality to deduce that the LHS is at least $(n+1)/2+(n-1)/2\cdot\sqrt[(n^2-n)/2]{\frac{A_1...A_{n-1}}{A_n^{n-1}}}$. Finally, kill the last beast with another AM-GM inequality.

This problem was used as problem 4 of the final exam of the 3rd TST of Taiwan 2005. (Only 1 person got it right.)

Solved with Amol Rama, Raymond Feng, Samuel Wang, and William Wang. Let $c_i=A_i/A_{i+1}$. The inequality we wish to prove reads $$n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}+\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le n+1.$$ We see that $$n\left(c_{n-1}^{\frac{n-1}{n^2}}c_{n-2}^{\frac{n-2}{n^2}}\cdots c_1^{\frac{1}{n^2}}\right)^{1/n^2}\le \frac{n-1}{n}c_{n-1}+\cdots+\frac{1}{n}c_1+\frac{n+1}{2},$$ and $$\left(1\cdot(2-c_1)\cdot(3-2c_2)\cdots(n-(n-1)c_{n-1})\right)^{1/n}\le \frac{n+1}{2}-\frac{1}{n}c_1-\cdots-\frac{n-1}{n}c_{n-1},$$ so adding gives the desired estimate. Remark: The solution is motivated by the $n=2$ case, which was solved with a similar AM-GM as above.

Note that $a_k=kA_k-(k-1)A_{k-1}$. Let $b_k=\frac{A_{k-1}}{A_k}$ for all $k$. Then, we have $$\begin{align*} n \sqrt[n]{\frac{G_n}{A_n}}+ \frac{g_n}{G_n} &= n \sqrt[n^2]{\frac{A_1A_2A_3\dots A_n}{A_n^n}} + \sqrt[n]{\frac{a_1}{A_1}\cdot \frac{a_2}{A_2}\cdot \ldots \cdot \frac{a_n}{A_n}} \\ &= n\sqrt[n^2]{\left(\frac{A_1}{A_n}\right)\left(\frac{A_2}{A_n}\right)\dots \left(\frac{A_{n-1}}{A_n}\right)} + \sqrt[n]{\left(\frac{2A_2-A_1}{A_2}\right)\left(\frac{3A_2-2A_2}{A_3}\right)\dots\left(\frac{nA_n-(n-1)A_{n-1}}{A_n}\right)} \\ &=n\sqrt[n^2]{1^{\frac{n(n+1)}{2}}b_1 b_2^2b_3^3\dots b_{n-1}^{n-1}}+\sqrt[n]{(1)(2-b_1)(3-2b_2)(4-3b_3)\dots(n-(n-1)b_{n-1})} \\ &\le n\cdot \frac{\frac{n(n+1)}{2}+b_1+2b_2+\dots + (n-1)b_{n-1}}{n^2} + \frac{\frac{n(n+1)}{2}-b_1-2b_2-\dots-(b-1)b_{n-1}}{n} \\ = n+1 \end{align*}$$ as desired.

Re-write the ineq as $n \cdot \sqrt[n^2]{\frac{A_1 \cdots A_{n-1}}{A_n^{n-1}}}+\sqrt[n]{\left(2-\frac{A_1}{A_2} \right) \cdots \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)} \le n+1$. Now from AM-GM we notice that we have that: $$\frac{n+1}{2}-\frac{\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n}=\frac{1+\left(2-\frac{A_1}{A_2} \right)+ \cdots + \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)}{n} \ge \sqrt[n]{\left(2-\frac{A_1}{A_2} \right) \cdots \left(n-\frac{(n-1)A_{n-1}}{A_n} \right)}$$ Now now all we need to prove after replacing this, is the following: $$\frac{n+1}{2}+\frac{\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n} \ge n \cdot \sqrt[n^2]{\frac{A_1 \cdots A_{n-1}}{A_n^{n-1}}}$$ But from AM-GM once again one has that: $$\frac{\frac{n(n+1)}{2}+\frac{A_1}{A_2}+\cdots+\frac{(n-1)A_{n-1}}{A_n}}{n^2} \ge \sqrt[n^2]{\frac{1^{\frac{n(n+1)}{2}} \cdot A_1 \cdots A_{n-1}}{A_n^{n-1}}}$$ (And thus eq case happens when $a_1=a_2=\cdots=a_n$) So simply rearranging this one gives the desired ineq, thus we are done .