Maximum value of $\lambda$ It is given that a,b,c are be of same sign and a,b,c are in Harmonic progression i.e. $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ and also
 $\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\geq \sqrt{\lambda \sqrt{\lambda \sqrt{\lambda.............\infty}}}$.
Find the greatest possible value of $\lambda$. 
How should I begin? I think we somehow need to find value of R.H.S. when equality holds but how to achieve that? Also using the result $b=\frac{2ac}{a+c}$ doesn't give any fruitful result.

Let $a=1/x,\; b=1/y, \;c=1/z.$ We have $y=(x+z)/2$. The LHS of the inequality is $$\frac {y+x}{2 y-x}+\frac  {y+z}{2 y-z}=\frac { z/2+ 3 x/2}{z}+\frac { x/2+ 3 z/2}{x}=1+\frac {3}{2} (x/z+z/x).$$ Since $x$ and $z$ (which are $1/a$ and $1/c$) have the same sign, $x/z$ and $z/x$ are positive, so $$x/z+z/x=(\;(x/z)^{1/2}- (z/x)^{1/2}\;)^2+2\geq 2.$$ Therefore $1+\frac {3}{2}(x/z+z/x)\geq 1+\frac{3}{2}(2)=4.$ Equality occurs here iff $a=b=c.$
