iterated square roots convergence I suspect that
$$
\forall (x, a) \in \mathbb{N}_*^2,\ \lim_{k \rightarrow \infty} \underbrace{\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{\dots \sqrt{x + a}}}}}}_k > \frac{x}{2}
$$
However, I have no  idea how to prove it.

Define
$$f(x)=\lim_{k \rightarrow \infty} \underbrace{\sqrt{x + \sqrt{x + \sqrt{x + \sqrt{\dots \sqrt{x + a}}}}}}_k$$
Assuming that $f(x)$ converges,
\begin{align}
f(x)&=\sqrt{x+f(x)}\\
\left\{f(x)\right\}^2&-f(x)-x=0\\
f(x)&=\frac{1+\sqrt{1+4x}}{2}
\end{align}
And then
$$f(x)\le\frac x2\quad(x\ge6)$$
