Edit: @2below is right :/

rafayaashary1 wrote: If it does not beat $r$ in both categories, it still has the property. Could you explain more about this step?

rafayaashary1 wrote:

This solution has a hole in the sense that it does not consider the case where \(x\) beats \(r\) when it comes to the food, but they have the same rating when it comes to the service, or vice versa. Well, we can easily see that there must exist at least one restaurant for which no other restaurant beats it when it comes to the food, call this the food restaurant. Using induction, assume the problem to be true for the case with \(n\) restaurants. For \(n+1\) restaurants, see that after removing this restaurant, by the induction hypothesis, there exists a restaurant among the \(n\) remaining restaurants satisfying this property, call it the overrated restaurant. Now if the overrated restaurant beats the food restaurant for the service, then we are done; the overrated restaurant satisfies the condition. Otherwise if the food restaurant beats the overrated restaurant when it comes to the service, then all the restaurants the overrated restaurant beats in terms of service are also beaten in service by the food restaurant, and for those restaurants which can't be compared to the overrated restaurant and the food restaurant in terms of service must be beaten by the food restaurant in terms of food, so we are done; the food restaurant satisfies the condition. Otherwise if the food restaurant and the overrated restaurant can't be compared in terms of service, then the food restaurant beats the overrated restaurant in terms of food. Consider all the restaurants which the overrated restaurant beats in service. If one of these restaurants beats the food restaurant in terms of service, then the overrated restaurant beats the food restaurant in terms of service, and we are done; the overrated restaurant satisfies the condition.. Otherwise either the food restaurant beats this restaurant in terms of service, or they cannot be compared in terms of service, meaning that the food restaurant beats this restaurant in terms of food. We are then done; the food restaurant satisfies this condition. Wait, I see a hole (in my proof)

We can think that restaurants are vertices and we have two colors (one for each category) of directed edges - edge from $x$ to $y$ tells that $x$ is better than $y$. Between each pair of vertices is at least one edge, one of them we can erase and we have tournament. In this graph theory interpretation it is Iran TST 2006 Problem 6: https://artofproblemsolving.com/community/c6h84260p487757

Name restaurants as $R_{1},\ldots, R_{n}$. Assign to restaurant $R_{k}$ two number $x_{k}$ and $y_{k}$. $x_{k}$ for rate of $R_{k}$ in tastiness and $y_{k}$ for it's rate in quality of service. If restaurants $R_{k}$ and $R_{l}$ aren't comparable in tastiness, then $x_{k}=x_{l}$ and $y_{k}=y_{l}$ if in quality of service. Second condition of problem is OK with comparing numbers. Lemma: Conditions of problem implies that $$ \exists i,j: \quad x_{i}=x_{j} \quad \Rightarrow \quad y_{i}\not=y_{j}$$$$ \exists i,j: \quad y_{i}=y_{j} \quad \Rightarrow \quad x_{i}\not=x_{j}$$It is so trivial. WLoG we can assume that $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$. Consider $x_{n}=x_{n-1}=\ldots = x_{m} > x_{m-1} \geq \ldots \geq x_{1}$. By the lemma it is clear that $$\forall m\leq i<j\leq n, \quad y_{i}\not= y_{j}$$Among $y_{m},y_{m+1},\ldots,y_{n}$, consider the biggest one, $y_{t}$. We know $y_{t}>y_{i}$ ($m \leq i \leq n$, $i \not= t$), thus every $R_{i}$ ($m \leq i \leq n$, $i \not= t$) is worse than $R_{t}$. Also we know $x_{t}>x_{j}$ ($1\leq j \leq m-1$), thus each $R_{j}$ ($1\leq j \leq m-1$) is worse than $R_{t}$. So $R_{t}$ is desired restaurant.$\blacksquare$

meysam1371 wrote: Name restaurants as $R_{1},\ldots, R_{n}$. Assign to restaurant $R_{k}$ two number $x_{k}$ and $y_{k}$. $x_{k}$ for rate of $R_{k}$ in tastiness and $y_{k}$ for it's rate in quality of service. If restaurants $R_{k}$ and $R_{l}$ aren't comparable in tastiness, then $x_{k}=x_{l}$ and $y_{k}=y_{l}$ if in quality of service. Second condition of problem is OK with comparing numbers. Lemma: Conditions of problem implies that $$ \exists i,j: \quad x_{i}=x_{j} \quad \Rightarrow \quad y_{i}\not=y_{j}$$$$ \exists i,j: \quad y_{i}=y_{j} \quad \Rightarrow \quad x_{i}\not=x_{j}$$It is so trivial. WLoG we can assume that $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$. Consider $x_{n}=x_{n-1}=\ldots = x_{m} > x_{m-1} \geq \ldots \geq x_{1}$. By the lemma it is clear that $$\forall m\leq i<j\leq n, \quad y_{i}\not= y_{j}$$Among $y_{m},y_{m+1},\ldots,y_{n}$, consider the biggest one, $y_{t}$. We know $y_{t}>y_{i}$ ($m \leq i \leq n$, $i \not= t$), thus every $R_{i}$ ($m \leq i \leq n$, $i \not= t$) is worse than $R_{t}$. Also we know $x_{t}>x_{j}$ ($1\leq j \leq m-1$), thus each $R_{j}$ ($1\leq j \leq m-1$) is worse than $R_{t}$. So $R_{t}$ is desired restaurant.$\blacksquare$ This doesn't work. Consider the case $x_1=x_2$, $x_1=x_3$, $x_1=x_4$, $x_2>x_3>x_4$ and $y_4>y_1>y_2>y_3$. According to you $R_1$ should be the one that beats everyone in at least one category, which is clearly not the case.

We'll proceed by complete induction. The base case $n=1$ is trivial. Suppose that it is true for every positive integer $m<n$. Well write \[ R_i<_a R_j \iff R_j \text{ beats } R_i \text{ in food.} \]\[ R_i<_b R_j \iff R_j \text{ beats } R_i \text{ in service.} \] Let $R_m$ be the restaurant that beats the most restaurants in at least one category. If it beats all other $n-1$ restaurants we are done, otherwise there must be a restaurant $R_p$ that beats $R_m$ in at least one category, let's say food, and it's not beaten by $R_m$ in the other category. Thus we write $R_m <_a R_p$. If $R_i<_a R_m$ for every restaurant $R_i$ that is beaten by $R_m$ then we get a contradiction because $R_i<_aR_p$ for all $i$, which would imply $R_p$ beats more restaurants than $R_m$. Thus, there must be some restaurants $R_j$ such that $R_j<_bR_m$ and $R_m$ does not beat $R_j$ in food. Consider the set $S$ of such restaurants. If one of them $R_j$ beats $R_p$ in service then we are done because $R_p<_bR_j$ but also $R_j<_bR_m$, thus $R_m$ also beats $R_p$, which is a contradiction. Now apply our induction hypothesis to $S \cup \{R_p\}$. If $R_p$ is the one that beats every other restaurant in $S \cup \{R_p\}$ then $R_p$ beats every restaurant that $R_m$ beats and also $R_m$ which is a contradiction. Thus, there is a restaurant $R_k \in S$ that beats everyone in $S \cup \{R_p\}$. Since $R_k$ cannot beat $R_p$ in service, in must be the case $R_k>_a R_p$. But $R_p>_aR_m$ and $R_m>_aR_j$ for all restaurants $R_j$ that $R_m$ beats that are not in $S$. Thus $R_k$ beats all restaurants that $R_m$ beats and also beats $R_m$ and $R_p$, which is more restaurants than $R_m$ beats, which is a contradiction. Thus $R_m$ must beat all other restaurants in at least one category.

Not sure if this is a fakesolve (got pointed out that my interpretation of the problem might be wrong)