# $\left \langle (A^{2}+B^{2})x,x \right \rangle\geq \left \langle (AB+BA)x,x \right \rangle$ if $A$ and $B$ are symmetric

Let be $$A$$ and $$B$$ two real matrices of $$n \times n$$. And $$\left \langle , \right \rangle$$ denotes the usual inner product in $$\mathbb{R}^{n}.$$

Prove that if $$A$$ and $$B$$ are symmetric then $$\forall x \in \mathbb{R}^{n}$$ it satisfies:

\begin{align*} \left \langle (A^{2}+B^{2})x,x \right \rangle\geq \left \langle (AB+BA)x,x \right \rangle \end{align*} Hint: Consider $$\left \langle (A-B)^{2}x,x \right \rangle$$

What I think I can do is to note that:

\begin{align*} \left \langle A^{2}+B^{2})x,x \right \rangle&=\left \langle A^2x,x \right \rangle + \left \langle B^2x,x \right \rangle\\\left \langle AB+BA)x,x \right \rangle&=\left \langle AB,x \right \rangle + \left \langle BA,x \right \rangle \end{align*}

And then try to prove in general that:

\begin{align*} \left \langle A^2x,x \right \rangle&\geq\left \langle ABx,x \right \rangle\\ \end{align*}

Neverthless, I don't know how to use the hint and the fact that the matrices are symmetric. Can you help me please? I would really appreciate it.

• $$(A-B)^2 = (A^2 + B^2) - (AB + BA)$$
• The eigenvalues of $$(A-B)^2$$ are nonnegative so $$\langle (A-B)^2 x, x \rangle \ge 0$$ for any $$x$$.
• $A$ and $B$ being symmetric implies that $A-B$ is symmetric, so $(A-B)^T(A-B) = (A-B)^2$, which implies the second point in the answer.