TITLE: Show that $\triangle ABC$ satisfying $B^2=AC$ and $2b=a+c$ is equilateral
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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.