67th Romanian Mathematical Olympiad

Clearly, $a_n\ge 1$, $n=1,2,3,\dots$. On the other hand, $$\begin{array}{rl}{a}_n-a_{n+1}& ={\int }_0^1\left(\frac{1+(f(x){)}^n}{1+(f(x){)}^{n+1}}-\frac{1+(f(x){)}^{n+1}}{1+(f(x){)}^{n+2}}\right)dx\\ & ={\int }_0^1\frac{(f(x){)}^n(1-f(x){)}^2}{(1+(f(x){)}^{n+1})(1+(f(x){)}^{n+2})}dx\ge 0,n=1,2,3,\dots ,\end{array}$$ so the sequence $(a_n{)}_{n\ge 1}$ is decreasing, and consequently convergent. Let $\ell ={\lim }_{n\to \infty }{a}_n$.

Without loss of generality we may and will assume that $f(0)=0$ and $f(1)=1$. Let $a=\inf \{x:0\le x\le 1,f(x)=1\}$. If $a=0$, then $f(x)=1,0<x\le 1$, so $a_n=1,n=1,2,3,\dots$, and $\ell =1$. If $a>0$, consider any positive $\epsilon <a$ and write $$\begin{array}{rl}0\le a_n-1& ={\int }_0^1\left(\frac{1+(f(x){)}^n}{1+(f(x){)}^{n+1}}-1\right)dx={\int }_0^1\frac{(f(x){)}^n(1-f(x))}{1+(f(x){)}^{n+1}}dx\\ & \le {\int }_0^1(f(x){)}^n(1-f(x))dx={\int }_0^a(f(x){)}^n(1-f(x))dx\\ & \le {\int }_0^a(f(x){)}^{n}dx={\int }_0^{a-\epsilon }(f(x){)}^{n}dx+{\int }_{a-\epsilon }^a(f(x){)}^{n}dx\\ & \le (a-\epsilon )(f(a-\epsilon ){)}^n+\epsilon ,n=1,2,3,\dots \end{array}$$ Since $0\le f(a-\epsilon )<1$, it follows that $(f(a-\epsilon ){)}^n\stackrel{n\to \infty }{\to }0$, so $0\le \ell -1\le \epsilon$, by the preceding; and since $\epsilon$ is arbitrary, we conclude that $\ell =1$.

Remark. Since the integrands form a sequence of measurable functions pointwise convergent to the constant function $1$, and each integrand is pointwise bounded from above by the constant (hence integrable) function $2$, the conclusion follows by Lebesgue's dominated convergence theorem.