[proofplan]
We introduce the shifted operator $S:=N-\lambda I_H$ and prove directly that $S$ is normal. The equality of norms then follows by expanding $\|Sx\|_H^2$ using the adjoint, commuting $S^*S$ to $SS^*$ by normality, and recognizing the result as $\|S^*x\|_H^2$. Finally, substituting $S^*=N^*-\overline{\lambda}I_H$ transfers any bounded-below estimate from $S$ to $S^*$ with the same constant.
[/proofplan]
[step:Show that the scalar shift of $N$ is normal]
Fix $\lambda\in\mathbb C$, and define the [bounded linear operator](/page/Bounded%20Linear%20Operator) $S\in\mathcal L(H)$ by
\begin{align*}
S:=N-\lambda I_H.
\end{align*}
For every $x,y\in H$,
\begin{align*}
(Sx,y)_H=((N-\lambda I_H)x,y)_H=(Nx,y)_H-\lambda(x,y)_H.
\end{align*}
Using the definition of the [Hilbert space](/page/Hilbert%20Space) adjoint and the convention that $(\cdot,\cdot)_H$ is linear in the first argument, this becomes
\begin{align*}
(Sx,y)_H=(x,N^*y)_H-(x,\overline{\lambda}y)_H=(x,(N^*-\overline{\lambda}I_H)y)_H.
\end{align*}
Therefore
\begin{align*}
S^*=N^*-\overline{\lambda}I_H.
\end{align*}
Now compute
\begin{align*}
S^*S=(N^*-\overline{\lambda}I_H)(N-\lambda I_H)
\end{align*}
and hence
\begin{align*}
S^*S=N^*N-\lambda N^*-\overline{\lambda}N+|\lambda|^2I_H.
\end{align*}
Similarly,
\begin{align*}
SS^*=(N-\lambda I_H)(N^*-\overline{\lambda}I_H)
\end{align*}
and therefore
\begin{align*}
SS^*=NN^*-\overline{\lambda}N-\lambda N^*+|\lambda|^2I_H.
\end{align*}
Since $N$ is normal, $N^*N=NN^*$. Thus $S^*S=SS^*$, so $S$ is normal.
[guided]
The point of introducing $S:=N-\lambda I_H$ is that the desired identity is exactly the standard norm identity for a normal operator and its adjoint, applied to this shifted operator. We first need to identify its adjoint. For $x,y\in H$, the operator $S$ satisfies
\begin{align*}
(Sx,y)_H=((N-\lambda I_H)x,y)_H=(Nx,y)_H-\lambda(x,y)_H.
\end{align*}
By the definition of $N^*$, we have $(Nx,y)_H=(x,N^*y)_H$. Because the Hilbert space [inner product](/page/Inner%20Product) is linear in the first argument and conjugate-linear in the second, the scalar term satisfies
\begin{align*}
\lambda(x,y)_H=(x,\overline{\lambda}y)_H.
\end{align*}
Therefore
\begin{align*}
(Sx,y)_H=(x,N^*y)_H-(x,\overline{\lambda}y)_H=(x,(N^*-\overline{\lambda}I_H)y)_H.
\end{align*}
Since this identity holds for all $x,y\in H$, the defining property of the adjoint gives
\begin{align*}
S^*=N^*-\overline{\lambda}I_H.
\end{align*}
We next verify normality of $S$ rather than assuming it. Multiplying the two factors gives
\begin{align*}
S^*S=(N^*-\overline{\lambda}I_H)(N-\lambda I_H).
\end{align*}
Expanding by bilinearity of operator composition and using $I_HN=N$ and $N^*I_H=N^*$ gives
\begin{align*}
S^*S=N^*N-\lambda N^*-\overline{\lambda}N+|\lambda|^2I_H.
\end{align*}
Likewise,
\begin{align*}
SS^*=(N-\lambda I_H)(N^*-\overline{\lambda}I_H)
\end{align*}
and expansion gives
\begin{align*}
SS^*=NN^*-\overline{\lambda}N-\lambda N^*+|\lambda|^2I_H.
\end{align*}
The only possible difference between these two expressions is the first term. Since $N$ is normal, $N^*N=NN^*$. Substituting this equality into the two expansions gives $S^*S=SS^*$, so the shifted operator $S=N-\lambda I_H$ is normal.
[/guided]
[/step]
[step:Convert normality of the shift into equality of the two norms]
Let $x\in H$. Since $S$ is normal, $S^*S=SS^*$. Using the defining property of the adjoint,
\begin{align*}
\|Sx\|_H^2=(Sx,Sx)_H=(S^*Sx,x)_H.
\end{align*}
By normality of $S$,
\begin{align*}
(S^*Sx,x)_H=(SS^*x,x)_H.
\end{align*}
Applying the defining property of the adjoint again, now to $S$ and the vector $S^*x$, gives
\begin{align*}
(SS^*x,x)_H=(S^*x,S^*x)_H=\|S^*x\|_H^2.
\end{align*}
Thus
\begin{align*}
\|Sx\|_H^2=\|S^*x\|_H^2.
\end{align*}
Both sides are non-negative [real numbers](/page/Real%20Numbers), so taking square roots yields
\begin{align*}
\|Sx\|_H=\|S^*x\|_H.
\end{align*}
Substituting $S=N-\lambda I_H$ and $S^*=N^*-\overline{\lambda}I_H$ gives
\begin{align*}
\|(N-\lambda I_H)x\|_H=\|(N^*-\overline{\lambda}I_H)x\|_H.
\end{align*}
[/step]
[step:Transfer the bounded-below estimate to the adjoint shift]
Assume that $N-\lambda I_H$ is bounded below with constant $c>0$, meaning that
\begin{align*}
\|(N-\lambda I_H)x\|_H\ge c\|x\|_H
\end{align*}
for every $x\in H$. By the norm equality already proved, for every $x\in H$,
\begin{align*}
\|(N^*-\overline{\lambda}I_H)x\|_H=\|(N-\lambda I_H)x\|_H.
\end{align*}
Combining the two displayed identities gives
\begin{align*}
\|(N^*-\overline{\lambda}I_H)x\|_H\ge c\|x\|_H
\end{align*}
for every $x\in H$. Since $(N-\lambda I_H)^*=N^*-\overline{\lambda}I_H$, this proves that $(N-\lambda I_H)^*$ is bounded below with the same constant $c$.
[/step]
