Problem 1
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that
(i) For all $x, y \in \mathbb{R}$,
$$
f(x)+f(y)+1 \geq f(x+y) \geq f(x)+f(y) \text {, }
$$
(ii) For all $x \in[0,1), f(0) \geq f(x)$,
(iii) $-f(-1)=f(1)=1$.

Find all such functions $f$.
Answer: $f(x)=\lfloor x\rfloor$, the largest integer that does not exceed $x$, is the only function.
Solution
Plug $y \rightarrow 1$ in (i):
$$
f(x)+f(1)+1 \geq f(x+1) \geq f(x)+f(1) \Longleftrightarrow f(x)+1 \leq f(x+1) \leq f(x)+2 .
$$

Now plug $y \rightarrow-1$ and $x \rightarrow x+1$ in (i):
$$
f(x+1)+f(-1)+1 \geq f(x) \geq f(x+1)+f(-1) \Longleftrightarrow f(x) \leq f(x+1) \leq f(x)+1 .
$$

Hence $f(x+1)=f(x)+1$ and we only need to define $f(x)$ on $[0,1)$. Note that $f(1)=$ $f(0)+1 \Longrightarrow f(0)=0$.
Condition (ii) states that $f(x) \leq 0$ in $[0,1)$.
Now plug $y \rightarrow 1-x$ in (i):
$$
f(x)+f(1-x)+1 \leq f(x+(1-x)) \leq f(x)+f(1-x) \Longrightarrow f(x)+f(1-x) \geq 0 .
$$

If $x \in(0,1)$ then $1-x \in(0,1)$ as well, so $f(x) \leq 0$ and $f(1-x) \leq 0$, which implies $f(x)+f(1-x) \leq 0$. Thus, $f(x)=f(1-x)=0$ for $x \in(0,1)$. This combined with $f(0)=0$ and $f(x+1)=f(x)+1$ proves that $f(x)=\lfloor x\rfloor$, which satisfies the problem conditions, as since
$$
x+y=\lfloor x\rfloor+\lfloor y\rfloor+\{x\}+\{y\} \text { and } 0 \leq\{x\}+\{y\}<2 \Longrightarrow\lfloor x\rfloor+\lfloor y\rfloor \leq x+y<\lfloor x\rfloor+\lfloor y\rfloor+2
$$
implies
$$
\lfloor x\rfloor+\lfloor y\rfloor+1 \geq\lfloor x+y\rfloor \geq\lfloor x\rfloor+\lfloor y\rfloor
$$