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We prove that the sequence \( \{ I_n \}_{n \in \mathbb{N}} \) is increasing using induction and the Cauchy-Schwarz inequality in a "smart" way.
Base case \( (n=0 )\): \[ \left( \int_{[a,b]} f(x) \, dx \right)^2 = \left( \int_{[a,b]} f(x) \frac{g(x)^{1/2}}{g(x)^{1/2}} \, dx \right)^2 \le \left( \int_{[a,b]} \frac{f(x)^2}{g(x)} \, dx \right) \left( \int_{[a,b]} g(x) \, dx \right) \]and thus \[ I_0 = \int_{[a,b]} f(x) \, dx \le \int_{[a,b]} \frac{f(x)^2}{g(x)} \, dx = I_1 . \]Note that actually the previous inequality is strict since it turns out that \[ I_1 - I_0 = \int_{[a,b]} \frac{(f(x)-g(x))^2}{g(x)} \, dx > 0 \]by the hypothesis on \(f\) and \(g\) (positivity and continuity).
Now write \[ \begin{split} I_{n-1} ^2= \left( \int_{[a,b]} \frac{ f(x)^{n}}{g(x)^{n-1}} \frac{f(x)^{n - \frac{n+1}{2}}}{f(x)^{n - \frac{n+1}{2}}} \frac{g(x)^{n-1 - \frac{n}{2}}}{g(x)^{n-1-\frac{n}{2}}} \, dx \right)^2 & \le \left( \int_{[a,b]} \frac{f(x)^{n+1}}{g(x)^n} \, dx \right) \left( \int_{[a,b]} \frac{f(x)^{n-1}}{g(x)^{n-2}} \, dx \right) \\ & =I_{n} I_{n-2} \end{split} \] by Cauchy-Schwarz inequality again. Now since \( I_{n-2} \le I_{n-1}\) by inductive hypothesis, we have \( I_{n-1} \le I_{n-1} ^2 / I_{n-2} \le I_n \). But using the base case we have \( I_0< I_1 ^2 / I_0 \le I_2 \) etc..., which implies that the inequalities are all strict, i.e. \[ I_0 < I_1 < I_2 < \dots < I_{n-1} < I_n . \] We have moreover \( I_n / I_{n-1} \ge I_{n-1} / I_{n-2}\) and \[ I_1 \left(\frac{I_1}{I_0}\right)^{n-1} \le I_1 \frac{I_1}{I_0} \frac{I_2}{I_1} \dots \frac{I_{n-1}}{I_{n-2}} \le I_n \]which implies \( I_n \to \infty \) when \( n \to \infty\).