$AC^2=\frac{BC^2+AB^2}{2}$ is equivalent to $\cos B=\frac{BC^2+AB^2-AC^2}{2BC \cdot AB}=\frac{AC^2}{2BC \cdot AB}$ We know that $\frac{AC^2}{2BC \cdot AB}=\frac{\sin^2 B}{2\sin A \sin C}$ Hence, $\cos B=\frac{\sin^2 B}{2\sin A \sin C}$ or equivalently $\cot B = \frac{\sin B}{2\sin A \sin C}$ So, we have to prove that $\left(\frac{\sin B}{2\sin A \sin C}\right)^2 \geq \frac{\cos A \cos C}{\sin A \sin C}$ The last one is equivalent to $\sin^2 B \geq 4 \sin A \sin C \cos A \cos C= \sin 2A \sin 2C$ Now, by AM-GM; $\sin 2A \sin 2C \leq \left(\frac{\sin 2A + \sin 2C}{2}\right)^2$ and by Jensen's inequality $\frac{\sin 2A + \sin 2C}{2} \leq \sin(A+C) = \sin B$ (Note that if one of $A$ and $C$ is obtuse, then the inequality is obvious since $\cot A \cot C<0$) So, we get $\sin 2A \sin 2C \leq \left(\frac{\sin 2A + \sin 2C}{2}\right)^2 \leq \sin^2 B$ hence, we are done.

Another way : Using the identity : $\boxed{\ \cot A=\frac {b^2+c^2-a^2}{4S}\ }$ , where $S$ denotes the area of $\triangle\ ABC$ , as well as the similar ones, the given inequality can be written as: $(c^2+a^2-b^2)^2\ge (b^2+c^2-a^2)(a^2+b^2-c^2)\iff$ $\iff b^4\ge b^4-(c^2-a^2)^2 \iff (c^2-a^2)^2\ge 0$ , what is obviously.

$\cot A=\frac{\cos A}{\sin A}$ etc so the inequality is equivalent to $\frac{\sin A\sin B}{\cos A\cos C}\ge\frac{\sin^2 B}{\cos^2 B}$. Now $b^2=a^2+b^2-2ac\cos B=2b^2-2ac\cos B\implies b^2=2ac\cos B$ so if we substitute $a=2R\sin A$ etc we get $\sin ^2 B=2\sin A\sin C\cos B$. Hence we aim to prove \begin{align*}&\frac{\sin A\sin B}{\cos A\cos C}\ge\frac{\sin^2 B}{\cos^2 B}\\ \iff & \frac{\sin A\sin B}{\cos A\cos C}\ge\frac{2\sin A\sin C\cos B}{\cos^2 B}\\ \iff &\cos B\ge 2\cos A\cos C\\ \iff &\frac{a^2+c^2-b^2}{2ac}\ge 2\left(\frac{b^2+c^2-a^2}{2bc}\right)\left(\frac{a^2+b^2-c^2}{2ab}\right)\\ \iff & b^2(a^2+c^2-b^2)\ge (b^2+c^2-a^2)(a^2+b^2-c^2)\end{align*} Substituting in $b^2=\frac{1}{2}(a^2+c^2)$ reduces the inequality to $(a^2+c^2)\ge (3c^2-a^2)(3a^2-c^2)$ which is true since $4(a^2-c^2)^2\ge 0$. Equality occurs when $a=c$, which then forces $\triangle ABC$ to be equilateral.