= Solution
Take a <normalization>[normalized] vector in the required common <operator domains>. Expanding the squared <norm> and using <self-adjointness> gives
$$
\|(A-itB)\psi\|^2=\langle\psi,(A+itB)(A-itB)\psi\rangle=\langle A^2\rangle+t^2\langle B^2\rangle-it\langle[A,B]\rangle.
$$
Thus its final term is $\hbar t\langle C\rangle$. Set $a=\|A\psi\|^2$, $b=\|B\psi\|^2$, $c=\langle C\rangle$. If $b>0$, minimizing the nonnegative quadratic at $t_*=-\hbar c/(2b)$ gives
$$
\boxed{4ab\geq\hbar^2c^2.}
$$
Equality holds precisely when the minimum <norm> vanishes, that is $(A-it_*B)\psi=0$. If $b=0$, however, then $B\psi=0$ and $c=0$, so equality is automatic <even> when $A\psi\ne0$. \b[The printed equality criterion needs this zero-vector exception.] For example $A=I$, $B=0$, $C=0$ saturates the inequality but admits no such real $t$. The exact general criterion is <linear dependence> of $A\psi,B\psi$ together with vanishing real part of their <inner product>, as in the <uncentered second-moment commutator inequality>.
Apply the same calculation to $A-\langle A\rangle I$, $B-\langle B\rangle I$, whose <commutator> is unchanged. This gives the <Robertson uncertainty principle>
$$
\boxed{4(\Delta A)^2(\Delta B)^2\geq\hbar^2\langle C\rangle^2.}
$$
If $\psi$ is an <eigenvector> of $A$, its <dispersion> is zero; finite <dispersion> of $B$ then forces $\boxed{\langle C\rangle=0}$.
In the <position representation in quantum mechanics>[position representation], $X\psi=x\psi$ and $P\psi=-i\hbar\psi'$. The <product rule> gives $[P,X]\psi=-i\hbar\psi$, hence $\boxed{[P,X]=-i\hbar I}$. Since $V(X)$ commutes with $X$,
$$
\boxed{[H,X]=\frac1{2m}\{P[P,X]+[P,X]P\}=-\frac{i\hbar}{m}P.}
$$
For an <energy eigenstate>, $\langle[H,X]\rangle=E\langle X\rangle-E\langle X\rangle=0$, so $\boxed{\langle P\rangle=0}$. For the oscillator $[H,P]=i\hbar m\omega^2X$, and the same argument gives $\boxed{\langle X\rangle=0}$. These <expectation> identities require the stated finite moments and <commutator> domains; formal position <eigenvectors> are not normalizable exceptions.
Apply the <Robertson uncertainty principle> to each angular-momentum pair:
$$
\boxed{4(\Delta L_1)^2(\Delta L_2)^2\geq\hbar^2\langle L_3\rangle^2.}
$$
If $\psi$ is an <eigenvector> of $L_3$, then $\Delta L_3=0$. The inequalities for $(L_2,L_3)$ and $(L_3,L_1)$ force $\boxed{\langle L_1\rangle=\langle L_2\rangle=0}$, assuming the relevant moments exist.
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