Does the inequality $ \sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2}$ have a name?

If we divide by $\sqrt 2$, the inequality becomes $$ \sqrt{\frac{a+b}{2}} \ge \frac{\sqrt{\vphantom b a} + \sqrt{b}}{2} $$ which is true because $\sqrt x$ is concave. (You can call this Jensen's inequality, though since we're only averaging $2$ points, it's just a special case of the definition of concavity.)

From AM-GM $$\frac{a+b}{2}\geq \sqrt{ab} \iff\\ a+b\geq \frac{a+b}{2}+2\sqrt{\frac{a}{2}\cdot\frac{b}{2}} \iff\\ a+b\geq \left(\sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}\right)^2 \overset{a,b\geq0}{\iff}\\ \sqrt{a+b}\geq \sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}$$ Which makes it a special case or corollary of it.

Squaring both sides we obtain

$$ \sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2} \iff a+b\ge \frac{a+b+2\sqrt{ab}}2\iff \frac {a+b}2\ge \sqrt{ab} $$

which is true by AM-GM and squaring again we obtain

$$\iff \frac{a^2+2ab+b^2}{4}\ge ab \iff a^2-2ab+b^2=(a-b)^2 \ge 0$$

which is a more foundamental result.