Given two densely defined unbounded self-adjoint strictly positive operator $A$ and $L$ in Hilbert space $H$ with domain $D(A) \subset D(L)$ and $\|Lx\| \leq \|Ax\|$ for all $x\in D(A)$, why do we also have $$\|(L+tI)^{1/2}x\| \leq \|(A+tI)^{1/2}x\|,$$ for all $t\geq 0$ and $D(A) = D((A+tI)^{1/2})$, $D(L) = D((L+tI)^{1/2})$ ?
Such a argument appears in the proposition 8.21 of this book.
Besides, I do not understand the duality argument in the proof as well.
Notice that $\|Lx\|^2 = \langle x, L^2 x\rangle$, and by assumption, this is $\le \|Ax\|^2=\langle x, A^2 x\rangle$. In other words, $L^2\le A^2$, and since the square root function is operator monotone, this implies that $L\le A$ and thus also $L+t\le A+t$. Now we obtain in the same way that $$ \|(L+t)^{1/2}x\|^2 = \langle x, (L+t) x\rangle\le \|(A+t)^{1/2}x\|^2 . $$ As for the domains, you must have misunderstood something. If $L$ is unbounded, then $D((L+t)^{1/2})\supsetneqq D(L)$. What is true is that you have the same inclusion as before for the domains of the square roots.