Denote $y_n=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx$ for all integer $n \geq 0$. Prove that $(y_n)_{n \geq 1}$ is an increasing and divergent sequence. Let $f,g:[0,1] \rightarrow (0,\infty)$ be two distinct, continuous functions such that $$\int_0^1 f(x)dx=\int_0^1 g(x)dx$$
Denote $$y_n=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx$$ for all integer $n \geq 0$. Prove that $(y_n)_{n \geq 1}$ is an increasing and divergent sequence.
Here is the source of the problem. 
My attempt: 
Note that $$y_n -y_{n-1}=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx-{\frac{f^{n}(x)}{g^{n-1}(x)}}dx = \int_0^1{\frac{f^{n+1}(x)-f^n(x)g(x)}{g^n(x)}}dx = \int_0^1{\frac{f^n(x)}{g^n(x)}(f(x)-g(x))}dx \geq A\int_0^1(f(x)-g(x)dx=0$$
where $A>0$.
Hence, $(y_n)_{n \geq 1}$ is increasing. However, I have trouble to show the sequence diverges. Can anyone give some hint?

There is an $c \in [0,1]$ such that $f(c)>g(c)$. Using the continuity we get that there is $\lambda >1$ and $[\alpha , \beta ] \subset [0,1]$ such that $\frac{f(x)}{g(x)} \ge \lambda $ on $[ \alpha , \beta ]$ . Then $y_n \ge \int_ {\alpha }^{\beta }f(x) \lambda ^n dx$, and so $\lim_{n \to \infty }y_n= \infty$.
