For any positive integer $n$: $f(x+1) \geq f(x+1-\frac1n) + \sqrt(\frac1n) \geq f(x+1-\frac2n) + 2\sqrt(\frac1n) \geq f(x+1-\frac3n) + 3\sqrt(\frac1n) \geq . . . \geq f(x)+\sqrt(n)$ so we can see that $f(x+1)$ has to be infinitely big for any $x$ since we can set $n$ to be as large as we want in the previous observation. Therefore no $f$ satisfies the condition for all $x,y$.

$(x,y)=(0,1/n)\implies f(1/n^2)\geq f(0)+1/n$ $(x,y)=(1/n^2,1/n)\implies f(2/n^2)\geq f(0)+2/n$ $(x,y)=(2/n^2,1/n)\implies f(3/n^2)\geq f(0)+3/n$ ...(iterate) $f(1)=f(n^2/n^2)\geq f(0)+n^2/n=f(0)+n$, absurd.

Let $P(x,y)$ denote the given assertion. $P(0,x): f(x^2) \ge x+f(0)$. $P(x^2, x): : f(2x^2) \ge f(x^2) + x\ge f(0) + 2x$. Claim: We have $f(n\cdot x^2) \ge f(0) + nx$ for all positive integers $n$ and nonnegative reals $x$. Proof: We induct on $n$, base cases $1,2$. If $f(k\cdot x^2) \ge f(0) + kx$, then \[P(kx^2, x): f((k+1)x^2) \ge f(k\cdot x^2) + x\ge f(0) + (k+1)x,\]as desired. $\square$ Let $Q(n,x)$ denote the assertion that \[f(n\cdot x^2) \ge f(0) + nx\] $Q(n,\sqrt{1/n}): f(1)\ge f(0) + n\sqrt{1/n} = f(0) + \sqrt{n}$. Setting $n$ sufficiently large gives a contradiction.

Let $P(x,y) : f(x+y^2) \geq f(x)+y$ . Let $k \in N$ $P(0,\frac{1}{k}) : f(\frac{1}{k^2}) \geq f(0) + \frac{1}{k}$ $P(\frac{1}{k^2}+\frac{1}{k}) : f(\frac{2}{k^2}) \geq f(\frac{1}{k^2})+\frac{1}{k} \geq f(0)+\frac{2}{k}$ .... $P(\frac{k-1}{k^2} , \frac{1}{k}) : f(1) \geq f(0)+\frac{k^2}{k}$ : In if $\lim_{k \to \infty }$ we can see is false ! . Now we don't have any function !