It is given that $f(x)$ is a function defined on $\mathbb{R}$, satisfying $f(1)=1$ and for any $x\in \mathbb{R}$, $$f(x+5)\geq f(x)+5,$$ $$f(x+1)\leq f(x)+1.$$ If $g(x)=f(x)+1-x$ then find $g(2002)$.
Here, $$f(x+5)\leq f(x+4) +1,$$ I didn't get any idea..
On
Rewriting the given inequalities in terms of $\,f(x)=g(x)+x-1\,$:
$$\require{cancel} \begin{align} f(x+5)\geq f(x) + 5 \quad&\iff\quad g(x+5)+\bcancel{x}+\cancel{5}- \xcancel{1} \ge g(x) + \bcancel{x} -\xcancel{1} + \cancel{5} \\ &\iff\quad g(x+5) \ge g(x) \\ f(x+1)\leq f(x)+1 \quad&\iff\quad g(x+1)+\bcancel{x}+\cancel{1}- \xcancel{1} \le g(x) + \bcancel{x} -\xcancel{1} + \cancel{1} \\ &\iff\quad g(x+1) \le g(x) \\ \end{align} $$
It follows that $\,\color{blue}{g(x)} \ge g(x+1) \ge g(x+2) \ge g(x+3) \ge g(x+4) \ge g(x+5) \ge \color{blue}{g(x)}\,$, so equalities must hold throughout, then $\,g(x+1)=g(x)\,$, and $\,g(2002)=g(2001)=\ldots =g(1)\,$.
You are in the right direction, just carry on:
$$f(x+5)\leq f(x+4) +1\leq f(x+3) +2\leq f(x+2) +3\leq f(x+1) +4\leq f(x) +5$$
Observe something fishy between the LHS and the RHS here? What can you say about $f(x)$ now?