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.

So, why don't we rewrite the HP condition in a nicer way, as 
$$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$
and let's multiply the numerator and denominator of the first fraction by $2/b$. We get
$$ \frac{a+b}{2a-b} = \frac{a(1/a+1/c)+2}{2a(1/a+1/c)-2} = \frac{3+a/c}{2a/c}=\frac{3c+a}{2a}$$
Similarly, the other one becomes $\frac{3a+c}{2c}$, so what we want to show reduces to
$$\frac{3c+a}{2a} + \frac{3a+c}{2c}\geq 4$$
which reduces to
$$\frac{c}{a}+\frac{a}{c}\geq 2$$
which follows directly from AM-GM.
