The mean value theorem and an arctangent inequality

Fix $t>0$. The function $\arctan$ is continuous on $[0,t]$ and differentiable on $(0,t)$ with derivative $\frac1{1+x^2}$. By the mean value theorem there exists $c\in(0,t)$ with
$$\arctan t-\arctan0=\frac1{1+c^2}(t-0),\qquad\text{i.e.}\qquad\arctan t=\frac t{1+c^2}.$$
Since $0<c<t$, we have $1+c^2<1+t^2$, so $\frac1{1+c^2}>\frac1{1+t^2}$. Multiplying by $t>0$,
$$\arctan t=\frac t{1+c^2}>\frac t{1+t^2}.$$

Remark. The inequality is *strict* precisely because $c<t$ strictly. The mean value theorem is the standard tool for turning information on $f'$ (here: $f'$ is decreasing in $|x|$) into information on $f$.