Quadratic Taylor majorant of a reciprocal quadratic ($\frac{x^2-4x+5}{4}-\frac{1}{1+x^2}=\frac{(x-1{)}^4}{4(1+x^2)}$)

The quadratic Taylor polynomial of $f(x)=1/(1+x^2)$ at $x=1$ is $p(x)=(x^2-4x+5)/4$. Multiplying $p-f$ by $4(1+x^2)$ gives $(x-1{)}^4$, proving the exact displayed identity. Thus $p\ge f$ for every real $x$, with equality only at one. In particular, a claim that $f>p$ to the right of one has the wrong sign. The Lagrange remainder also proves this direction there, since $f^{\prime \prime \prime }(x)=24x(1-x^2)/(1+x^2{)}^4<0$ for $x>1$. The exact identity is stronger: it controls the sign globally and shows the fourth-order contact at the expansion point.