Find all $f$ which satisfies $ f:\mathbb{R}_{\geq0} \rightarrow \mathbb{R}, f(x+y^2) \geq f(x)+y $

Question

$$ f:\mathbb{R}_{\geq0} \rightarrow \mathbb{R}, f(x+y^2) \geq f(x)+y $$ First I thought was $f(x)=c\sqrt{x}$. Putting to the original F.E.: $$ c\sqrt{x+y^2} \geq c\sqrt{x}+y \\ \ \\ \text{let } c \geq 0. \\ c^2(x+y^2) \geq c^2x + cy\sqrt{x}+y^2 \\ c^2y^2 \geq y(c\sqrt{x}+y) \\ \text{if }y = 0: 0\geq0 \rightarrow \text{No Problem.} \\ \text{if }y \neq 0: (c^2-1)y \geq c\sqrt{x} \\ \ \\ \therefore c \leq 0. \Rightarrow \text{if } c = 0 \Rightarrow 0 \geq y \Rightarrow \otimes. \\ \therefore c < 0. $$

Answer 1

Let $P(x,y)$ denote the statement $f(x+y^2) \geq f(x)+y$.

There is no such function. Note that for any $n\in\mathbb{N}$ and $1\leq k\leq n$ we have $$P\left(x+\frac{k}{n},\frac{1}{\sqrt{n}}\right):\ f\left(x+\frac{k}{n}\right)\geq f\left(x+\frac{k-1}{n}\right)+\frac{1}{\sqrt{n}}.$$ Adding up these inequalities we get $f(x+1)\geq f(x)+\sqrt{n}$. For $x$ fixed and $n$ sufficiently large this a contradiction.