I have an increasing sequence $(x_n)$ with $0 < x_1 < 1$ and a recursive estimate $$x_n^2 \leq x_1^2 + x_{n-1}^4(1-x_1^2)$$ I want to prove $$\sup_n x_n^2 \leq \frac{x_1^2}{1-x_1^2}$$
Trying induction:
For $n = 1$, since we have for the denominator $0 <1-x_1^2 <1$, the inequality holds.
Assume the inequality holds for $n=k$, giving us $$\sup_k x_k^2 \leq \frac{x_k^2}{1-x_k^2}$$
Now if I consider $$x_{k+1}^2\leq x_1^2 + x^4_{k}(1-x_1^2)$$ Now since I have a order 4 term, it makes it difficult to get back the previous form. Since the RHS of the inequality looks like the sum of a geometric series $$S_n = x_1^2 +x_1^4 +x_1^6+...$$ I think there should be a way to write the order 4 term as a larger sum that I can have as a power of this GP but I can't figure it out. Any pointers appreciated
We prove by induction.
For $n = 1$, since we have for the denominator $0 <1-x_1^2 <1$, the inequality holds.
Assume the inequality holds for $n=k$, giving us $$\sup_{1\le i \le k} x_i^2 \leq \color{red}{\frac{x_1^2}{1-x_1^2}}$$
For $n = k+1$, we have $$ \begin{align} x_{k+1}^2 \le x_1^2 + x_k^4 (1-x_1^2) &\le x_1^2 + \left( \frac{x_1^2}{1-x_1^2} \right)^2 (1-x_1^2) \\ &= x_1^2 + \frac{x_1^4}{1-x_1^2} \\ &= x_1^2 \left(1 + \frac{x_1^2}{1-x_1^2} \right) \\ &= \frac{x_1^2}{1-x_1^2} \\ \end{align} $$ So, the statement holds true for $n = k+1$.
Q.E.D