I am looking to prove that
$$ \frac{(x-a)^2}{(b+\sqrt{x})^2} \leq c^2 \implies x \leq 2a^2 + 4c^2 + 2b^2 $$
for nonnegative $x, a, b$ and obviously $c$.
My idea is to obtain a bound based on the positive root of a quadratic equation. Indeed, the inequality implies that
$$ x^2 - 2 a^2 -4c^2 b^2-4c^2 x \leq 0 $$
which implies that
$$x \leq \frac{4c^2+ \sqrt{16c^4 + 8a^2 + 32c^2 b^2}}{2} $$
but although this bound looks like what I was hoping to get, it's not quite it.
Can you please provide a hint on how to get the target bound? Thank you.
You can use the binomial formulas: $$(x-a)^2-c^2(b+\sqrt{x})^2\le 0$$ so we get $$(x-a-c(b+\sqrt{x}))(x-a+c(b+\sqrt{x}))\le 0$$ Substituting $$\sqrt{x}=t$$ we get an quadratic inequality $$t^2-ct-a-bc\le 0$$