Note that \[ \frac{a}{b^2+1} = a\left(1-\frac{b^2}{b^2+1}\right) = a -\frac{ab^2}{b^2+1}\ge a - \frac{ab}{2}, \]using $b^2+1\ge 2b$. Using $a+b+c+d=4$, it thus suffices to show, \[ 4\ge ab+bc+cd+da = (a+c)(b+d). \]But this is trivial: \[ (a+c)(b+d)\le \left(\frac{a+b+c+d}{2}\right)^2 = 4, \]by AM-GM.

Related to the tangent line trick, find constants $A$ and $B$ such that $\frac{1}{x^2 + 1} \geq Ax+B$ for $x\in (0,4)$, i.e. $(Ax+B)(x^2 + 1) - 1$ has $1$ as a double root - only $B = 1$, $A = -\frac{1}{2}$ will work. So we reduce to $a(1 - \frac{b}{2}) + b(1-\frac{c}{2}) + c(1-\frac{d}{2}) + d(1-\frac{a}{2}) \geq 4$, i.e. $(a+c)(b+d) \leq 4$ which holds since $(a+c)(b+d) \leq \frac{((a+c)+(b+d))^2}{4} = 4$.