is IRN,TWN-Iran ,Taiwan?

Yup, it is

The answer is all $f$ such that $f(x)\equiv c|x|$ for $c\le 2$. Let $P(x,y)$ denote assertion in the question. $P(x,0) \rightarrow f(0)=0$ $P(0,y) \rightarrow f(-y)\le f(y)\le f(-y) \to f(y)=f(-y)$ $P(y,y) \rightarrow f(y)\ge 0$ for all $y$. By induction on $n$, we can show that $f(ny)=nf(y)\forall n\in \mathbb{N}, y\in \mathbb{R}$. Claim: If $f(x),f(y)>0$ and $x,y>0$, then $f(x+y)=f(x)+f(y)$ Proof: The main idea is that there exists $cx-dy\in [0,1)$ such that $f(cx-dy)$ is unbounded. We will induct on $c+d$ to show that $f(cx-dy)=cf(x)+df(y)\forall c,d\in \mathbb{N}_0$. Base Case: $0\le c,d\le 1$, clear. Inductive Step: Assume it's true for $(c,d)$ and $(c,d-1)$, then it's true for $(c,d+1)$ by $P(cx+dy,y)$ because $f(cx+(d-1)y) = cf(x)+(d-1)f(y) \ne cf(x)+(d+1)f(y)$. If $\frac xy=\frac mn\in \mathbb{Q}^+$ then $f(x)=mf(\frac xm), f(y)=nf(\frac yn)$ and $f(x+y)=f(x)+f(y)$ clearly holds. Otherwise there exists arbitrarily large $c,d$ st $cx-dy \in (0,1)$, done. Therefore, if $x>y$, then $f(x)=f(x-y)+f(y)$ if $f(x-y), f(y)>0$. Claim: For all $x$ such that $f(x)>0$, $\frac{f(x)}{x}$ is constant. Proof: same trick; if $\frac{f(x)}{x} < \frac{f(y)}{y}, x,y>0$ then there exists $c,d \in \mathbb{N}$ such that $cy-dx \in (0,1)$ and $f(cy-dx) > 2$. Therefore, if $f(x)>0, f(y)=0$ then either $f(x-y)=f(x)$ or $f(x+y)=f(x)$, so $y=0$. Thus, $f(x)\equiv c|x|$ for some $c\le 2$.

2022 Iran Taiwan Friendly Math Competition/2 wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that: $f(x) < 2$ for all $x \in (0,1)$ For all real numbers $x,y$, we have \[ \max \{ f(x + y), f(x - y) \} = f(x) + f(y) \] Proposed by Navid Safaei The answer is $\boxed{f(x) = c|x|}$ for all $x \in \mathbb{R}$ for any $c \in [0,2]$. This clearly satisfies the first condition and to see this satisfy the second condition, see that If $x,y$ have the same sign, then \[ \max \{ f(x+y), f(x - y) \} = \max \{ c|x + y|, c|x - y| \} = c | x + y| = c(|x| + |y|) = c|x| + c|y| = f(x) + f(y) \]If $x,y$ have opposite sign, then \[ \max \{ f(x + y), f(x - y) \} = \max \{ c |x + y|, c | x- y| \} = c|x - y| = c(|x| + |y|) = c|x| + c|y| = f(x) + f(y) \]Now, we'll prove that these are the only solutions to the functional equation above. Let $P(x,y)$ be the assertion of $x$ and $y$ to the functional equation \[ \max \{ f(x + y), f(x - y) \} = f(x) + f(y) \]Obviously, $f \equiv 0$ work. Suppose now that $f$ is not the zero function. $P(0,0)$ gives us $f(0) = 0$, and $P(0,x)$ gives us $f(x) \ge f(-x)$ for all $x \in \mathbb{R}$ and thus we conclude that $f(x) \ge f(-x) \ge f(-(-x)) = f(x)$, i.e. $f(x) = f(-x)$ for all $x \in \mathbb{R}$. Claim 01. $f(nx) = |n|f(x)$ for any $n \in \mathbb{Z}$ and $x \in \mathbb{R}$. Proof. It suffices to consider the case where $x > 0$. We'll consider two cases: If $f(a) = 0$, then we will prove the following statement for all $n \in \mathbb{N}$ by induction: $f((n + 1)a) \le 0$ and $f(ia) = 0$ for all $i \le n$, $i \in \mathbb{N}$. For $n = 1$, $P(a,a)$ gives us $f(2a) \le 0$ as desired. Now, suppose that the statement hold for $n = k$, i.e. we have $f(ia) = 0$ for all $i \le k$ and $f((k + 1)a) \le 0$. Then $P((k + 1)a, a)$ gives us \[ \max \{ f((k + 2)a), 0 \} = \max \{ f((k + 2)a), f(ka) \} = f((k + 1)a ) + f(a) = f((k + 1)a) \]However, we know that \[ \max \{ f((k + 2)a), 0 \} \ge 0 \ge f((k + 1)a) \]by hypothesis, and equality must hold. This implies that $f((k + 2)a) \le 0$, and $f((k + 1)a) = 0$ as desired. This is enough to prove that $f(na) = nf(a)$ for all $n \in \mathbb{N}$ -- and therefore $f(n) = |n|f(a)$ for all $n \in \mathbb{Z}$. If $f(a) \not= 0$, then we will prove by induction that $f(na) = nf(a)$ for all $n \in \mathbb{N}$. This is clearly true for $n = 1$. $P(a,a)$ gives us \[ \max \{ f(2a), 0 \} = 2f(a) \not= 0 \implies f(2a) = 2f(a) \]Now, suppose that $f(ka) = kf(a)$ and $f((k - 1)a) = (k - 1)f(a)$ for some $k \ge 2$. Then, we have \[ P(ka, a) : \max \{ f((k + 1)a), (k - 1)f(a) \} = (k + 1)f(a) \]If $(k - 1)f(a) \ge f((k + 1)a)$, then we have $f(a) = 0$, a contradiction. Therefore, we must have $f((k + 1)a) = (k + 1)f(a)$, which proves what we wanted, i.e. $f(n) = nf(a)$ for all $n \in \mathbb{N}$ -- and thus $f(n) = |n|f(a)$ for all $n \in \mathbb{Z}$. Claim 02. $f(ma) = |m|f(a)$ for all $m \in \mathbb{Q}$ and $a \in \mathbb{R}$. Proof. It suffices to prove this for $m > 0$. Write $m = \frac{p}{q}$ for some $p,q \in \mathbb{N}$ such that $\gcd(p,q) = 1$. Then, we have \[ q f \left( \frac{p}{q} a \right) \stackrel{\text{Claim 01}}{=} f(pa) = pf(a) \implies f \left( \frac{p}{q} a \right) = \frac{p}{q} f(a) \]as desired. Claim 03. $f(x) > 0$ for all $x > 0$. Proof. $P(x,x)$ gives us $\max \{ 2f(x), 0 \} = 2f(x) \implies 2f(x) \ge 0 \implies f(x) \ge 0$ for all $x > 0$. Now, suppose that there exists $a \not= 0$ such that $f(a) = 0$. WLOG $a$ is positive. We have \[ f(x + y) \le \max \{ f(x + y), f(x - y ) \} = f(x) + f(y) \]However, for any $x \in \mathbb{R}$, we know that there exists $n \in \mathbb{Q}$ such that $x - na \in (0,1)$; and thus \[ f(x) \le f(x - na) + f(na) < 2 + |n|f(a) = 2 \]However, if there exists $y \not= 0$ such that $f(y) \not= 0$ then, we could easily take $N \in \mathbb{Q}$ such that $f(Nq) > 2$, a contradiction. Therefore, $f(x) \not= 0$ for all $x > 0$, i.e. $f(x) > 0$ for all $x > 0$. Claim 04. $f(x+y) = f(x) + f(y)$ for all $x,y > 0$. Proof. Suppose there exists $a > b > 0$ such that $f(a - b) = f(a) + f(b)$. Call the pair $(m,n) \in \mathbb{N}_0^2$ nice if $f(ma - nb) = mf(a) + nf(b)$. Clearly, $(x,0)$ and $(0,x)$ is nice for any $x \in \mathbb{N}_0$. Now, suppose that $(m,n) \in \mathbb{N}^2$ is nice for all $m + n \le k$, for $k \ge 2$. We'll now prove that $(m,n) \in \mathbb{N}^2$ is nice for all $m + n = k + 1$. Take any $m,n \in \mathbb{N}$ such that $m + n = k$, then $P(ma - nb, a)$ gives us \begin{align*} \max \{ f((m+1)a - nb), f((m - 1)a - nb) \} &= f(ma - nb) + f(a) \\ \max \{ f((m+1)a - nb), (m - 1)f(a) + nf(b) \} &= (m + 1)f(a) + nf(b) > (m - 1)f(a) + nf(b) \end{align*}and thus $(m+1,n)$ is nice. Similarly, $P(ma - nb, b)$ gives us \begin{align*} \max \{ f(ma - (n - 1)b), f(ma - (n + 1)b) \} &= f(ma - nb) + f(b) \\ \max \{ mf(a) + (n - 1)f(b), f(ma - (n + 1)b \} &= mf(a) + (n + 1)f(b) > mf(a) + (n - 1)f(b) \end{align*}and thus $(m,n + 1)$ is nice. Therefore, we conclude that any $(x,y)$ such that $x + y = k + 1$ is nice, i.e. all $(x,y) \in \mathbb{N}_0^2$ is nice. Now, there exists an unbounded increasing sequence $a_1, \dots, a_k, \dots $ and $b_1, \dots, b_k, \dots$ of natural numbers such that $a_i a - b_i b \in (0,1)$ for all $i$. However, by the first condition, we have \[ +\infty = \lim_{i \to \infty} a_i f(a) + b_if(b) = \lim_{i \to \infty} f(a_i a - b_ib) < 2 \]which is a contradiction. Now, we know that $f$ is additive in $\mathbb{R}^+$ and $f(x) > 0$ for all $x > 0$. This implies $f$ is linear in $\mathbb{R}^+$ and by Claim 02, we conclude that $f(x) = cx$ for $x > 0$ for some $c \in \mathbb{R}^+$. The first condition and $f(x) > 0$ implies that $c \ge 0$ and \[ c = \lim_{x \to 1} cx \stackrel{cx < 2 \ \forall x \in (0,1)}{\le} 2 \]Therefore, $f(x) = cx$ for all $x \ge 0$ for some $c \in [0,2]$. Using the fact that $f(-x) = f(x)$ for all $x \in \mathbb{R}$, we conclude that $f(x) = c|x|$ for all $x \in \mathbb{R}$ for some $c \in [0,2]$, as desired.