A problem from the Shortlist of the Romanian Mathematics Olympiad

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 $(b-a) \int^b_a f(x)g(x) dx > \int^b_a f(x) dx \int^b_a g(x) dx$.

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.