Let $0\leq a_1\leq a_2\leq\hdots\leq a_{n-1}\leq a_n$ such that $a_1+a_2+\hdots+a_n=1$. Prove for any non-negative numbers $x_1,x_2,\hdots,x_n$, $y_1,y_2,\hdots,y_n$ the inequality $$\left(\sum_{i=1}^n a_ix_i - \prod_{i=1}^n x_i^{a_i}\right) \left(\sum_{i=1}^n a_iy_i - \prod_{i=1}^n y_i^{a_i}\right) \leq a_n^2\left(n\sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i} - \sum_{i=1}^n\sqrt{x_i} \sum_{i=1}^n\sqrt{y_i}\right)^2.$$

Sorry for double posting; I just want to have a LaTeX'ed version that can be added into the contest collections.

MarkBcc168 wrote: Let $0\leq a_1\leq a_2\leq\hdots\leq a_{n-1}\leq a_n$ such that $a_1+a_2+\hdots+a_n=1$. Prove for any non-negative numbers $x_1,x_2,\hdots,x_n$, $y_1,y_2,\hdots,y_n$ the inequality $$\left(\sum_{i=1}^n a_ix_i - \prod_{i=1}^n x_i^{a_i}\right) \left(\sum_{i=1}^n a_iy_i - \prod_{i=1}^n y_i^{a_i}\right) \leq a_n^2\left(n\sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i} - \sum_{i=1}^n\sqrt{x_i} \sum_{i=1}^n\sqrt{y_i}\right)^2.$$ Thanks.

It's hard. The official solution is about 2.5 pages (P1-3 adds up to less than 1.5 pages). As far as I know, nobody successfully worked it out during the contest.

BUMPPPPP

minecraftfaq wrote: It's hard. The official solution is about 2.5 pages (P1-3 adds up to less than 1.5 pages). As far as I know, nobody successfully worked it out during the contest. Do you have the official solution?

Bump!!!!

Claim 1: (Dealing with weighted sums on left hand side) $$\sum_{i=1}^n a_ix_i - \prod_{i=1}^n x_i^{a_i} \le a_n \left( \sum_{i=1}^n x_i - n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} \right)$$ Proof. Rearrange as $$ \sum_{i=1}^{n-1} (a_n-a_i)x_i+ \prod_{i=1}^n x_i^{a_i} \ge n a_n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} .$$By weighted AM-GM inequality, $$ \sum_{i=1}^{n-1} (a_n-a_i)x_i+ \prod_{i=1}^n x_i^{a_i} \ge n a_n \left( \prod_{i=1}^{n-1} x_i^{a_n-a_i} \cdot \prod_{i=1}^n x_i^{a_i} \right) ^{1/(\sum_{i=1}^{n-1} (a_n-a_i) + \sum_{i=1}^n a_i)} =n a_n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} \blacksquare$$ Using Claim 1 (which holds similarly for $y_i$), it suffices to prove $$ \left( \sum_{i=1}^n x_i - n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} \right) \left( \sum_{i=1}^n y_i - n \left( \prod_{i=1}^n y_i \right)^{\frac{1}{n}} \right) \le \left(n\sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i} - \sum_{i=1}^n\sqrt{x_i} \sum_{i=1}^n\sqrt{y_i}\right)^2 .$$ Claim 2: (Separating $x_i$ and $y_i$ on right hand side) $$ \left( n \sum_{i=1}^n x_i - \left( \sum_{i=1}^n \sqrt{x_i} \right)^2 \right) \left( n \sum_{i=1}^n y_i - \left( \sum_{i=1}^n \sqrt{y_i} \right)^2 \right) \le \left(n\sqrt{\sum_{i=1}^n x_i\sum_{i=1}^n y_i} - \sum_{i=1}^n\sqrt{x_i} \sum_{i=1}^n\sqrt{y_i}\right)^2 .$$ Proof. By Cauchy-Schwarz, both brackets on left hand side are non-negative. Then it's a direct consequence of Aczel's inequality for $n=2$ (alternatively, Cauchy-Schwarz). $\blacksquare$ Using Claim 2, we are left to prove the following $$ \sum_{i=1}^n x_i - n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} \le n \sum_{i=1}^n x_i - \left( \sum_{i=1}^n \sqrt{x_i} \right)^2 .$$which can be rearranged as $$ (n-1) \sum_{i=1}^n x_i + n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} \ge \left( \sum_{i=1}^n \sqrt{x_i} \right)^2$$ By Surányi's Inequality and Cauchy-Schwarz, \begin{align*} (n-1) \sum_{i=1}^n x_i + n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{n}} \right) \left( \sum_{i=1}^n x_i^{\frac{n-1}{n}} \right) \\ & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{2}} \right)^2. \end{align*}We are done.

Rickyminer wrote: By Surányi's Inequality and Cauchy-Schwarz, \begin{align*} (n-1) \sum_{i=1}^n x_i + n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{n}} \right) \left( \sum_{i=1}^n x_i^{\frac{n-1}{n}} \right) \\ & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{2}} \right)^2. \end{align*} We don't need to use Suranyi's Inequality here. We will prove that for any $n$ positive reals $a_1, a_2, \cdots, a_n$, the following inequality holds. $$(n-1) \sum_{i=1}^{n} a_i^2 + n \sqrt[n]{\prod_{i=1}^{n} a_i^2} \ge \left( \sum_{i=1}^{n} a_i \right)^2$$Let $x_i = \log a_i$ for all $1 \le i \le n$, and $f = \exp$. Let $\bar{x} = \dfrac{x_1 + x_2 + \cdots + x_n}{n}$. The inequality is equivalent with the following. $$(n-1)(f(2x_1) + f(2x_2) + \cdots + f(2x_n)) + nf(2\bar{x}) \ge f(2x_1) + f(2x_2) + \cdots + f(2x_n) + 2\sum_{1 \le i < j \le n}f(x_i+x_j)$$We use Karamata here, as $f$ is concave up. Without loss of generality, assume that $x_1 \ge x_2 \ge \cdots \ge x_n$. It suffices to prove that $$(\underbrace{2x_1, \cdots, 2x_1}_{n-2 \text{ times}}, \underbrace{2x_2, \cdots, 2x_2}_{n-2 \text{ times}}, \cdots, \underbrace{2x_n, \cdots, 2x_n}_{n-2 \text{ times}}, \underbrace{2\bar{x}, \cdots, 2\bar{x}}_{n \text{ times}}) \text{ majorizes } (x_1 + x_2, x_1 + x_2, \cdots, x_{n-1} + x_n, x_{n-1} + x_n)$$Since a sequence $\textbf{a}$ majorizes a sequence $\textbf{b}$ if and only if $\sum_{i=1}^{n} (|a_i-x| - |b_i-x| ) \ge 0$ for any real $x$, we only need to prove that for any real $r$, $$2(n-2) \sum_{i=1}^{n} | x_i - r| + 2n | \bar{x} - r| \ge 4 \sum_{1 \le i < j \le n}|x_i+x_j - 2r|$$By shifting the sequence, we only need to consider $r = 0$, which is equivalent to a well known inequality, $$(n-2)\sum_{i=1}^{n}|x_i| + |\sum_{i=1}^{n} x_i| \ge 2 \sum_{1 \le i < j \le n} |x_i + x_j|$$

Rickyminer wrote: By Surányi's Inequality and Cauchy-Schwarz, \begin{align*} (n-1) \sum_{i=1}^n x_i + n \left( \prod_{i=1}^n x_i \right)^{\frac{1}{n}} & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{n}} \right) \left( \sum_{i=1}^n x_i^{\frac{n-1}{n}} \right) \\ & \ge \left( \sum_{i=1}^n x_i^{\frac{1}{2}} \right)^2. \end{align*}We are done. Absolute beauty! Wow. Bravo. The best inequality problem I've ever seen.