Consider a quantum system in a state $$| \psi \rangle$$, and let $$A$$ and $$B$$ be Hermitian operators representing a pair of observables. One can choose to perform a measurement of $$A$$ or a measurement of $$B$$ on the system. Generally, the outcomes of these measurements cannot be predicted with certainty. The outcomes are statistical in nature and characterized by the respective expectation values $$\langle A \rangle$$ and $$\langle B \rangle$$. Hereinafter, $\langle \cdot \rangle = \langle \psi | \cdot | \psi \rangle \,.$ The corresponding uncertainties are defined as $\sigma_A = \sqrt{ \langle \big( A - \langle A \rangle \big)^2 \rangle } = \sqrt{\langle A^2 \rangle - \langle A \rangle^2} \,,$ $\sigma_B = \sqrt{ \langle \big( B - \langle B \rangle \big)^2 \rangle } = \sqrt{\langle B^2 \rangle - \langle B \rangle^2} \,.$ The Robertson-Schrödinger uncertainty relation sates that $\sigma_A^2 \sigma_B^2 \ge \langle \tfrac{i}{2} [A,B] \rangle^2 + \Big( \langle \tfrac{1}{2} \{ A, B \} \rangle - \langle A \rangle \langle B \rangle \Big)^2 \,,$ where $[A,B] = A B - B A \,,$ $\{ A, B \} = A B + B A \,.$ As the operators $$A$$ and $$B$$ are Hermitian, so are the operators $$i[A,B]$$ and $$\{A,B\}$$, implying that the right-hand side of the Robertson-Schrödinger relation is non-negative.
Proof: The relation follows from the Cauchy-Schwarz inequality: $\big| \langle u | v \rangle \big|^2 \le \langle u | u \rangle \langle v | v \rangle \,,$ for any two states $$| u \rangle$$ and $$| v \rangle$$. Let us take $| u \rangle = \big( A - \langle A \rangle \big) | \psi \rangle \,,$ $| v \rangle = \big( B - \langle B \rangle \big) | \psi \rangle \,.$ Then, $\langle u | u \rangle = \langle \big( A - \langle A \rangle \big)^2 \rangle = \sigma_A^2 \,,$ $\langle v | v \rangle = \langle \big( B - \langle B \rangle \big)^2 \rangle = \sigma_B^2 \,,$ $\langle u | v \rangle = \langle \big( A - \langle A \rangle \big) \big( B - \langle B \rangle \big) \rangle = \langle A B \rangle - \langle A \rangle \langle B \rangle \,.$ Substituting these expressions into the Cauchy-Schwarz inequality, we obtain $\sigma_A^2 \sigma_B^2 \ge \big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2 \,.$ It remains to show that $$\big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2$$ can be rewritten as the right-hand side of the Robertson-Schrödinger uncertainty relation. To this end, we define two Hermitian operators $$C_+$$ and $$C_-$$ as $C_+ \equiv \{A,B\} = A B + B A \,,$ $C_- \equiv i [A,B] = i (A B - B A) \,.$ Then, $A B = \tfrac{1}{2} C_+ - \tfrac{i}{2} C_- \,,$ $B A = \tfrac{1}{2} C_+ + \tfrac{i}{2} C_- \,,$ and $\big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2 = \left| \langle \tfrac{1}{2} C_+ \rangle - \langle A \rangle \langle B \rangle - i \langle \tfrac{1}{2} C_- \rangle \right|^2 \,.$ Finally, since $$\langle C_+ \rangle$$, $$\langle C_- \rangle$$, and $$\langle A \rangle \langle B \rangle$$ are all real, we obtain $\big| \langle A B \rangle - \langle A \rangle \langle B \rangle \big|^2 = \Big( \langle \tfrac{1}{2} C_+ \rangle - \langle A \rangle \langle B \rangle \Big)^2 + \langle \tfrac{1}{2} C_- \rangle^2 \,,$ which concludes the proof.