I was looking at problem A3 on the Putnam 2017 set:
Let $a$ and $b$ be real numbers with $a<b$, and let $f$ and $g$ be continuous functions from $[a,b]$ to $(0, \infty)$ such that $\int_a^b f(x)\,dx = \int_a^b g(x)\,dx$ but $f \neq g$.
For every positive integer $n$, define
$$ I_n = \int_a^b \frac{(f(x))^{n+1}}{(g(x))^n}\,dx. $$ Show that $I_1, I_2, I_3, \dots$ is an increasing sequence with $\lim_{n \to \infty} I_n = \infty$.
and attempted the following solution.
Let $X$ be the subset of $[a,b]$ such that for all $x$ in $X$, $f(x) > g(x)$. Note that from the conditions on $f$ and $g$ such a set is guaranteed to have infinite size and have least one continuous range within it.
The above integral can be rewritten $$ \int_X \frac{(f(x))^{n+1}}{(g(x))^n}\,dx + \int_{[a,b]-X} \frac{(f(x))^{n+1}}{(g(x))^n}\,dx$$
We then take any $x_0$ in $X$, and observe that $$\lim_{n\rightarrow\infty} \frac{(f(x_0))^{n+1}}{(g(x_0))^n} = \lim_{n\rightarrow\infty} \left( \frac{f(x_0)}{g(x_0)} \right)^n f(x_0) = \infty$$ because $f(x_0) / g(x_0) > 1$ and $f(x_0) > 0$. (Also both are constant in respect to $n$).
Since every point in the left integral goes to infinity with $n$, the integral itself will as well. We can also see that the right integral must be positive with the fact that the ranges of $f$ and $g$ are positive. Therefore the original integral, $I_n$, also goes to infinity with $n$.
I am suspicious that there is an error somewhere. I have not seen this solution when looking through other solutions online, plus this solution would also be sufficient for $(f(x))^n / (g(x))^n$ (without the $+1$ in the top exponent), which seems odd. Is this a valid solution?