When both of $x, y$ are nonpositive, the quantity $\frac{x}{y^2+1}+\frac{y}{x^2+1}$ is obviously nonpositive as well, and the inequality holds in this case. Now consider when one of $x, y$ is nonpositive and the other is positive. WLOG $y$ is nonpositive and $x$ is positive. Then we have \[\frac{x}{y^2+1}+\frac{y}{x^2+1}\leq \frac{x}{x+1}+\frac{y}{x^2+1}=1-\frac{1}{x+1}+\frac{y}{x^2+1}<1,\]as desired. All that's left is to prove the case where both $x, y$ are positive. From the given inequalities we have $x^4\geq y^2\geq x\implies x\geq 1$, and similarly $y\geq 1$. Let $x+y=a$ and $xy=b$. It's clear that $a\geq 2$. Adding the two given conditions we have \[(x+y)^2-2xy\geq x+y\implies \frac{1}{2}(a^2-a)\geq b.\] It suffices to prove that \[x(x^2+1)+y(y^2+1)\leq (x^2+1)(y^2+1),\]or \begin{align*}x^3+y^3+x+y\leq x^2y^2+x^2+y^2+1 \implies a^3-3ab+a\leq b^2+a^2-2b+1. \end{align*}Putting everything on one side, we need to prove \[(a^2+1)(1-a)+b(3a-2)+b^2\geq 0.\].

The above post did a great job of narrowing the problem down to the case where x and y are positive. From now on, assume x and y are both positive. By taking the 2nd derivative of the given expression, we can see that it is a convex function in x. Thus, it remains to check that the inequality holds when x is at its minimum and maximum. Combining the first 2 conditions gets $x^4 \ge x$ and $y^4 \ge y$, or $x,y \ge 1$. The maximum and minimum of x are achieved at $x=y^2$ and $x=\sqrt{y}$, respectively. Plugging $x=y^2$ in and simplifying, it remains to show that $y^6+y^3+y^2+y \le y^6+y^4+y^2+1 \implies (y^3-1)(y-1) \ge 0$, clearly true Plugging $x=\sqrt{y}$ in and simplifying, it remains to show that $y^3 + y^{\frac{3}{2}} + y + y^{\frac{1}{2}} \le y^3+y^2+y+1 \implies (y\sqrt{y}-1)(\sqrt{y}-1) \ge 0$, clearly true