Trivial? Rewrite the condition as $(\sqrt{a}-\sqrt{bc})^2+b+c-bc=0$ so we are done.

Yes, it is easy inequality Also $2\sqrt{abc}=a+b+c \geq 2\sqrt{a(b+c)} \to bc \geq b+c$

Even more by the book - the equation $x^2 - 2\sqrt{bc}x + b + c = 0$ must have a real root (here $x = \sqrt{a}$) and the discriminant $D$ is $\frac{D}{4} = bc - (b+c)$.