Can anyone share a solution or the official solution? The problem seems tough(of course, it is an A9)

Note that the LHS is $2 - 2\sum_{i = 1}^{n} \min(x_i, y_i).$ Therefore, we just need: $$2 \sum_{i = 1}^{n} \min(x_i, y_i) \ge a + b,$$ where $a = \min \frac{x_i}{y_i}$ and $b = \min \frac{y_i}{x_i}.$ Let $A$ be the sum of the $x_i$'s so that $x_i < y_i$ and $B$ be the sum of the $y_i$'s so that $y_i \le x_i.$ We know that $1 = \sum x_i \le A + \frac{B}{b}$ and $1 = \sum y_i \le B + \frac{A}{a}.$ Hence, we have that: $$A + \frac{B}{b} \ge 1, \qquad (1)$$ and $$\frac{A}{a} + B \ge 1, \qquad (2).$$ By $(\frac1{a} -1) \cdot (1) + (\frac1{b}-1) \cdot (2),$ we obtain: $$(A+B)(\frac1{ab} - 1) \ge \frac1{a} + \frac1{b} - 2.$$ It therefore suffices to show that: $$\frac{\frac1{a} + \frac1{b} - 2}{\frac1{ab} - 1} \ge \frac{a+b}{2},$$ which is equivalent to: $$ab(a+b) + a + b \ge 4ab$$ upon cross-multiplying and rearranging. We'll show that this is true whenever $0 \le a, b \le 1$, which implies the result. Consider the function $f(a, b) = ab(a+b) + a + b - 4ab$. Note that for it to be maximal on $[0, 1]^2$, we need either $\{a, b\} \cap \{0, 1\} \neq \emptyset$ or that the partial derivatives of $f$ w.r.t $a, b$ are both $0$. In the former case, we have $f(a, 1) = (b-1)^2 \ge 0$ and $f(a, 0) = b \ge 0$, so we're done here. In the latter case, we have $2ab + b^2 + 1 - 4b = 2ab + a^2 + 1 - 4a = 0$ which implies $a(4-a) = b(4-b).$ As $a, b \in [0, 1]$ this gives $a = b$ and so $f(a, b) = 2a^2 + 2a - 4a^2 = 2a(a-1)^2 \ge 0.$ $\square$

Wow! Really complicated solution! I couldnt possibly solve this one!

In fact, we will prove something sharper: each of the terms $ \displaystyle \min_{1\leq i\leq n}\frac{x_i}{y_i}$ and $ \displaystyle \min_{1\leq i\leq n}\frac{y_i}{x_i}$ can be replaced with a convex combination of those of them which are less than $ 1$ (see $ (2)$ below). Some notations first. Let $$ \displaystyle I=\{i \mid 1\le i\le n, x_i\le y_i\}; J=\{i \mid 1\le i\le n, y_i<x_i\}.$$Note that $ I$ cannot be empty and $ J$ can be empty only when $ x_i=y_i,i=1,2,\dots,n$ in which case the inequality is trivial. Thus, we can assume both $ I,J$ are not empty sets. Denote $ \displaystyle k_i:=\frac{x_i}{y_i}, \forall i\in I$ and $ \displaystyle \ell_j:=\frac{y_j}{x_j}, \forall j\in J.$ Let $ k:=\min\{k_i\mid i\in I\}, \ell := \min\{\ell_j\mid j\in J\}.$ Clearly $ \displaystyle k= \min_{1\leq i\leq n}\frac{x_i}{y_i}, \ell=\min_{1\leq i\leq n} \frac{y_i}{x_i} .$ Assume $ i\in I.$ We have $$ \displaystyle |x_i-y_i|=y_i-x_i=(x_i+y_i)-2x_i= (x_i+y_i)-2k_iy_i.$$In the same way, for any $ j\in J$ $$ \displaystyle |x_j-y_j|=x_j-y_j=(x_j+y_j)-2y_j= (x_j+y_j)-2\ell_j x_j.$$Thus, it yields \begin{align*} \displaystyle \sum_{i=1}^n|x_i-y_i|&= \sum_{i\in I}\big((x_i+y_i)-2k_iy_i\big)+\sum_{j\in J}\big((x_j+y_j)-2\ell_j x_j\big)\\ &= \sum_{i=1}^n(x_i+y_i) -2\sum_{i\in I}k_iy_i-2\sum_{j\in J}\ell_jx_j\\ &=2 -2\sum_{i\in I}k_iy_i-2\sum_{j\in J}\ell_jx_j. \end{align*}Thus, in order to prove the original inequality, it is enough to prove $$ \displaystyle 2\sum_{i\in I}k_iy_i+2\sum_{j\in J}\ell_jx_j \ge k+\ell \qquad(1)$$We will prove something even stronger. It holds $$ \displaystyle 2\sum_{i\in I}y_ik_i+2\sum_{j\in J}x_j\ell_j \ge \sum_{i\in I}y_i^*k_i +\sum_{j\in J}x_j^*\ell_j\qquad (2)$$where $ \displaystyle y_i^*:=\frac{y_i}{\sum_{i\in I}y_i},\forall i\in I$ and $ \displaystyle x_j^*:=\frac{x_j}{\sum_{j\in J}x_j},\forall j\in J.$ Let us first see why $ (2)$ is sharper than $ (1).$ By definition, $ \displaystyle \sum_{i\in I} y_i^*=1, \sum_{j\in J} x_j^*=1 .$ Thus, the RHS of $ (2)$ consists of two convex combinations of the points $ k_i,i\in I,$ respectively $ \ell_j, j\in J$ and apparently $ \displaystyle \sum_{i\in I} y_i^*k_i\ge k, \sum_{j\in J} x_j^*\ell_j\ge \ell .$ In order to prove $ (2)$ we use $$ \displaystyle \sum_{i\in I} y_ik_i +\sum_{j\in J}x_j=1$$$$ \displaystyle \sum_{j\in J}x_j\ell_j + \sum_{i\in I}y_i=1.$$Denote $$ \displaystyle y:= \sum_{i\in I} y_ik_i\,;\, x:=\sum_{j\in J}x_j\ell_j.$$Hence, $ \displaystyle \sum_{j\in J}x_j =1-y, \sum_{i\in I} y_i=1-x$. Putting it back into $ (2),$ we need to prove $$ \displaystyle 2(x+y)\ge \frac{x}{1-y}+\frac{y}{1-x}\qquad (3)$$for any $ x,y$ satisfying $ 0<x,y<1, x+y\le 1.$ Let $ t=1-x-y, t\ge 0.$ The inequality $ (3)$ is equivalent to $$ \displaystyle 2(1-t)\ge \frac{x}{x+t}+\frac{y}{y+t}$$$$ \displaystyle 2t\le \frac{t}{x+t}+\frac{t}{y+t} \qquad (4)$$But, since $x+t=1-y\leq 1$ and $y+t=1-x\leq 1$ we indeed get $ \frac{t}{x+t}+\frac{t}{y+t} \ge t+t=2t$ which proves $ (4)$ and hence $ (2)$ holds. Remark. I am very glad that in recent years inequalities like this have been preferred to the standard ones. You can find some more thoughts on Olympic inequalities, as well as motivation, in my blog.

A comment on the solution in #3: $a^2b+ab^2+a+b=(a+b)(ab+1) \geq (a+b)^2 \geq 4ab$ since $(a-1)(b-1) \geq 0$. Otherwise, the idea for the two sums $A$ and $B$ and the bounds (1) and (2) are very nice and clever.

Very nice problem