A problem from the Shortlist of the Romanian Mathematics Olympiad Prove that $$\int_0^x \left(1+\frac{t}{1!}+\frac{t^2}{2!}+...+\frac{t^{2n+1}}{(2n+1)!} \right)\cdot \frac{1}{1+t^2} dt <\frac{\arctan x}{x} \int_0^x e^t dt$$ for all $x>0$. (it is not stated in the question, but I suppose that $n$ is just a positive integer)
I tried to use the fact that $e^t= \sum\limits_{k=0}^\infty \frac{t^k}{k !}$. This gave me that $$1+\frac{t}{1!}+\frac{t^2}{2!}+...+\frac{t^{2n+1}}{(2n+1)!}< e^t, \forall t\in [0,x]$$for some arbitrarily fixed $x$.
This lead to $$\int_0^x \left(1+\frac{t}{1!}+\frac{t^2}{2!}+...+\frac{t^{2n+1}}{(2n+1)!} \right)\cdot \frac{1}{1+t^2} dt < \int_0^x \frac{e^t}{1+t^2} dt.$$From here I tried to apply IBP, but it doesn't seem to get closer to the required inequality. 
 A: Alternative solution:
It suffices to prove that for $x > 0$, 
$$\tfrac{\arctan x}{x} (\mathrm{e}^x - 1) - \int_0^x \tfrac{\mathrm{e}^t}{1+t^2} \mathrm{d}t > 0.$$
Denote LHS by $f(x)$. We have
$$f'(x) = \frac{x\mathrm{e}^x - \mathrm{e}^x + 1}{x^2}\Big(\arctan x - \frac{x}{1+x^2}\Big).$$
Let $g(x) = \arctan x - \frac{x}{1+x^2}$. We have
$g'(x) = \frac{2x^2}{(x^2+1)^2} > 0$ for $x > 0$.
Also, $g(0) = 0$. Thus, we have $g(x) > 0$ for $x > 0$.
Let $g_1(x) = x\mathrm{e}^x - \mathrm{e}^x + 1$. 
We have $g_1'(x) = x\mathrm{e}^x > 0$ for $x>0$. Also, $g_1(0)=0$.
Thus, we have $g_1(x) > 0$ for $x > 0$. 
Thus, $f'(x) > 0$ for $x > 0$. Also, $\lim_{x\to 0+} f(x) = 0$.
Thus, we have $f(x) > 0$ for $x > 0$. We are done.
A: One can apply the Integral Chebyshev inequality to $f(t) = \frac{1}{1+t^2}$ and $g(t) = e^{t}$. Since $f$ and $g$ are of opposite monotonicity on $[0, x]$ this gives
$$
 \int_0^x f(t) g(t) \, dt < \frac 1x \int_0^x f(t) \, dt \int_0^x g(t) \, dt
$$
for $x > 0$. Strict inequality holds because both functions are not constant.
For a proof of that integral inequality see for example Given two increasing continuous functions $f,g$ prove that $(b-a) \int^b_a f(x)g(x) dx > \int^b_a f(x) dx \int^b_a g(x) dx$.
