Show that $\sqrt{9+a} \geq \sqrt{a} + \frac{2\sqrt{2}}{\sqrt{1 + a}}$

Question

I encountered this recreational problem:

Show that $\sqrt{9+a} \geq \sqrt{a} + \frac{2\sqrt{2}}{\sqrt{1 + a}} \quad \forall a \geq 0$. I also have to assess in what cases the equality holds. I am new to problem solving and tried some algebraic manipulations but don't seem to get far.

Answer 1

all is positive and we can square the inequality $$9+a>a+\frac{8}{1+a}+\frac{4\sqrt{2}\sqrt{a}}{\sqrt{1+a}}$$ it is equivalent to $$\frac{1+9a}{1+a}>\frac{4\sqrt{2}\sqrt{a}}{\sqrt{1+a}}$$ this is equivalent to $$(1+81a^2+18a)(1+a)>32a(1+a^2+2a)$$ in this is $$(1+a)(7a-1)^2>0$$ is true.