I am trying to prove that the following sequence of function $$p_{n+1}(x) = p_n(x) + \frac{x^2-p_n^2(x)}{2}, \; (n \geq 0)$$ with $p_0=0$ is monotonically increasing on the interval $[-1,1]$.
My attempt was to use induction on $n$. However, I am stuck with proceeding a set of inequalities in my inductive step. I wonder if there is a clever way to deduce that $x^2-p_{k+1}^2(x) \geq 0$ using my inductive hypothesis. I wrote down my attempt below.
Let $T(n)$ be the statement that $p_n(x) \leq p_{n+1}(x)$ $\forall x \in [-1,1]$.
(Base Case) If $n=1$, we have $p_1(x) = \frac{x^2}{2}$ and \begin{equation} p_2(x) = \frac{x^2}{2} + \frac{1}{2}\left(x^2-\frac{x^4}{4}\right) = x^2 - \frac{x^4}{8} \end{equation} It is clear that \begin{equation} p_1(x) - p_2(x) = \frac{x^2}{2}-x^2 + \frac{x^4}{8} = -\frac{x^2}{2} + \frac{x^4}{8} = \frac{x^2}{8}\left( x^2-4 \right) \leq 0 \end{equation} $\forall x \in [-1,1]$ since $x^2 < 4$. Hence, $T(1)$ holds.
(Inductive Step) Suppose $T(m)$ holds for all $m=1,\ldots,k-1$. We deduce that $T(k)$ holds, i.e., \begin{equation} p_k(x) \leq p_{k+1}(x) \iff 0 \leq p_{k+1}(x)-p_k(x) \end{equation} Note that since $x \in [-1,1]$, $p_k(x)^2 \leq p_k(x)$. Then, \begin{align} p_{k+1}(x)-p_k(x) = \frac{1}{2}(x^2-p_k^2(x)) \geq \frac{1}{2}(x^2 - p_k(x)) \geq \cdots \end{align} I have some difficulties figuring out how to proceed from here. I appreciate any hint or help. Thank you.