It was easy to see that for an equilateral triangle maximum value of $\angle B$ can be $\angle B=60^\circ$ as for other triangles $\angle C$ or $\angle A$ would have the largest angle, depending on the common difference of the sides.
But how do I prove it using trigonometry, geometry or even calculus
I tried taking sides as $a-d,\; a,\; a+d$ and applying $Law \;of\; sines\;$ but couldn't get the result.
Let the sides be $b-d, b, b+d$. To form a triangle, one needs $d<b$ and $b+d<b+b-d$, which gives $0\le\frac{d}{b}<\frac{1}{2}$
Then apply the law of cosines:
$$b^2=(b+d)^2+(b-d)^2-2(b+d)(b-d)\cos B$$
Long story short, you will get (unless I made some mistakes):
$$\cos B= \frac{3}{2(1-(d/b)^2)}-1$$
$B$ is maximum when $\cos B$ is minimum, which happens when $\frac{d}{b}=0$, so $d=0$ and the triangle is equilateral.