The given inequality can be rewritten as
$$(f(x+y))^2 \le f((x-y)^2).$$

Take $y = 1-x$. Then $1 = 1^2 = (f(1))^2 \le f((2x-1)^2)$. As $(2x-1)^2$ obtains all non-negative real values, we can conclude that $1 \le f(z)$ for every non-negative real number $z$.
Now take $x \ge -\frac{1}{2}$ and $y = 1+x$. Then $(f(2x+1))^2 \le f(1) = 1$. This implies $f(2x+1) \le 1$ because all values of $f$ are non-negative by the above. As $2x+1$ obtains all non-negative real values, we can deduce that $f(z) \le 1$ for every non-negative real number $z$.

Consequently, $f(z) = 1$ for every non-negative real number $z$. This function satisfies all conditions of the problem.