Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that$$f(x+y)\leq f(x^2+y)$$for all $x,y$.
solution
Let $0 \le d \le \frac{1}{4}$. Then we can find $x_1$ such that $x_1^2-x_1=d$ and $x_2$ such that $x_2^2-x_2=-d$. It follows that $f(x+d) \le f(x) \le f(x+d)$ for all $x$ and hence $f(x)=f(x+d)$ for all $x$. So $f$ is periodic with each period in $\left[0,\frac{1}{4}\right]$ and hence constant.
question 1
although it is obvious that the function is constant form the periodicity but I wanna know how to prove it rigorously.
question 2
Instead of saying that the function is periodic can I say every point is a local minimum and therefore it implies that the function is constant.
If every number on $[0,\frac 1 4]$ is a period of $f$ then every real number is a period of $f$ becasue any integer multiple of a period is always a period. If $x <y$ the $f(x)=f(x+(y-x))$ because $y-x$ is a period so $f(x)=f(y)$. This proves that $f$ is a constant.
Q2: You are probably mistaking $f(x+d) \leq f(x) \leq f(x+d)$ for $f(x-d) \leq f(x) \leq f(x+d)$. We don't get a local minimum from these inequalities. Besides the argument using local minimum doesn't work because $f$ is not given to be a continuous function.