To prove the given inequality Question:-
If $a,b,c$ are positive real numbers which are in H.P. show that $$\dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b} \ge 4$$

Attempt at a solution:-
I tried it by AM-GM inequality, but got stuck at a step. My attempt was as follows:-
$$\dfrac{ \dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b}}{2} \ge \sqrt{\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b}}$$
Evaluating the right hand side of the inequality,
$$\left(\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b} \right) = \left( \dfrac{b^2+\left(a+c \right)b+ac}{b^2-2b\left(a+c \right)+4ac}\right)$$
Now, as $a,b,c$ are in H.P., hence we get the relation $$b=\dfrac{2ac}{a+c}$$ On substituting this result in the equation, we get
$$\dfrac{10 a^2c^2 + 3ac(a^2+c^2)}{4a^2c^2}=\dfrac{5}{2}+\dfrac{3(a^2+c^2)}{4ac}$$
Now from the above calculations, what I observed is that, that for the proof to be valid, the condition $\dfrac{3(a^2+c^2)}{4ac}=\dfrac{3}{2}$ should be satisfied. The final conclusion to which I arrived was $a=c$. Now although the numbers would still be in H.P. but it doesn't prove the result for the numbers which are in H.P. and are not equal.

Any other approach to the proof is invited too.

The theorem you are asked to prove is not true in general. The additional condition needed is that $a \geq b$.  And then you can do the proof without AM-GM.
Assuming $(a,b,c)$ are in H.P. in the order $$a = \frac{1}{x}, b = \frac{1}{x+d}, c = \frac{1}{x+2d}$$
then the LHS is
$$
\frac{4x^2+8dx+6d^2}{x^2+2dx}
$$
Consider the case of $x^2+2dx > 0$:
$$
\frac{4x^2+8dx+6d^2}{x^2+2dx}-4 = \frac{1}{x^2+2d}\left[4x^2+8dx+6d^2 - 4(x^2+2dx) \right] = \frac{6d^2}{x^2+2d} \geq =0
$$
Now consider the case where  $x^2+2dx < 0$ and $d\neq 0$:
$$
\frac{4x^2+8dx+6d^2}{x^2+2dx}-4 =  \frac{6d^2}{x^2+2d} < 0
$$
So a good counterexample would be $x=3, d=-1$, giving $a=\frac13, b = \frac12, c=1$.
If you add the condition that $a\geq b$ then you find that forces $x^2 + 2dx \geq 0$ and the statement then is true.
