If $A = \int_Xf\ d\mu$, then $\sqrt{1+A^2} \ \le \ \int_x\sqrt{x+f^2}\ d\mu \ \le \ 1+A $

Question

Suppose $(X,M,\mu)$ is a measue space with $\mu(X)=1$ and $f:X\rightarrow[0,\infty)$ is measureble. Let $$A = \int_Xf\ d\mu$$ Show that $$ \sqrt{1+A^2} \ \le \ \int_X\sqrt{x+f^2}\ d\mu \ \le \ 1+A $$

The first equality seems obvious but I don't know how to officially prove and I'm completely stuck with the second equality. Any help would be appreciated!