Yes, this is true. I right $(x,y)=x'y=y'x$ for an inner product of (column) vectors $x,y$. We have $a=Ax=By$ for certain $x,y$ and also $a=(A+B)z$ (since the image of the operator $A+B$ is an orthogonal complement to the kernel of $A+B$ and thus contains the image of $A$.) We have $(Ax,x)(Az,z)\geqslant (Ax,z)^2=(a,z)^2$ (that is Cauchy-Bunyakovsky-Schwarz for the vectors $\sqrt{A}x,\sqrt{A}z$), therefore either $(a,x)=(a,z)=0$ or $(Az,z)\geqslant (a,z)^2/(a,x)$. Analogously $(Bz,z)\geqslant (b,z)^2/(a,y)$. Summing up we get $(a,z)\geqslant (a,z)^2(1/(a,x)+1/(a,y))$, $(a,x)(a,y)\geqslant (a,z)((a,x)+(a,y))$ as desired (if $(a,x)=(a,z)=0$ this is also so.)

But $(A^-a,a)=(A^-a,Ax)=(AA^-a,x)=(AA^-Ax,x)=(Ax,x)=(a,x)$ and so on, thus we have proved exactly what you ask for.