A $\triangle ABC$ satisfies the conditions below: $$B^2=AC \qquad 2b=a+c$$ Show that it's an equilateral one, where $a=|BC|$, $b=|AC|$, $c=|AB|$ and $B=\angle ABC$, $A=\angle BAC$, $C=\angle ACB$.
What I have done:
When trying to solve these question, I want to find a solution in all triangles which satisfy $2b=a+c$. It means that point $B$ can be regarded as a point in an ellipse.
Then I try to prove that for all points $B$ in that ellipse, we have $B^2\leq AC$. When proving it I think I can get the equal conditions and then show that $AC=B^2$ if and only if $A=B=C$ or $AC=0$.
Then I meet a terrible problem, meaning that I need to show ($x=A$):
$$ \frac{4-5\cos x}{5-4\cos x}\leq\cos\frac{(2\pi-x)-\sqrt{x(4\pi-3x)}}{2}=-\cos \frac{x+\sqrt{x(4\pi -3x)}}{2}，x\in[0,\frac{\pi}{3}]\\ $$
I have never seen such a hard problem before. I try to prove the inequality by segment amplification and minification, derivative, and many other ways but all failed.
Can you help me or give me some hints on this problems?
I am assuming $A\ge C$ without loss of generality. Therefore $A=Br$ and $C=\frac Br$ for some $r\ge1$, by the first equation. From the second equation, note that \begin{align} &2b=a+c\\ \implies &\dfrac{2b}{2R} = \dfrac{a}{2R}+\dfrac{c}{2R}\\ \implies &2\sin B = \sin A+\sin C&&(\text{By sine rule})\\ \implies &4\cos\dfrac B2\sin\dfrac B2 = 2\sin\left(\dfrac{A+C}{2}\right)\cos\left(\dfrac{A-C}{2}\right)\\ \implies &2\sin\dfrac B2=\cos\left(\dfrac{A-C}{2}\right)&&\left(\because\cos\frac B2 = \sin\left(\dfrac{A+C}{2}\right)\right)\\ \implies &2=\dfrac{\cos\left(\dfrac{A-C}{2}\right)}{\cos\left(\dfrac{A+C}{2}\right)}\\ \implies &3=\cot\left(\dfrac{A}{2}\right)\cot\left(\dfrac{C}{2}\right)&&(\text{By componendo and dividendo})\\ \implies &\tan\left(\frac A2\right)\tan\left(\frac C2\right)=\frac13\\ \implies &\tan\left(\dfrac{Br}{2}\right)\tan\left(\dfrac{B}{2r}\right)=\frac13\tag1\label{1} \end{align} Also note that \begin{align} &A+B+C=\pi\\ \implies &Br+B+\dfrac Br=\pi\\ \implies &B = \dfrac{\pi r}{r^2+r+1}\tag2\label{2} \end{align} Using \eqref{2} in \eqref{1}, we get, \begin{equation} \tan\left(\dfrac{\frac\pi2\times r^2}{r^2+r+1}\right)\tan\left(\dfrac{\frac\pi2}{r^2+r+1}\right)=\frac13 \end{equation} Now, since LHS is a decreasing function of $r$ for all $r\ge 1$ and equality holds at $r=1$, hence $r=1$ and we're done.