Question

If positive numbers $x,y,z$ are in $A.P.$ then the minimum value of $\frac{x+y}{2y-x}+\frac{y+z}{2y-z}$ is equal to -

Answer 1

We have,

$x,y,z$ are in an A.P.

Then,

$y=\frac{x+z}{2}$

$2y=x+z$

Let $f\left(x\right)=\frac{x+y}{2y-x}+\frac{y+z}{2y-z}$

Therefore,

$f\left(x\right)=\frac{x+\frac{x+z}{2}}{x+z-x}+\frac{\frac{x+z}{2}+z}{x+z-z}$

$\Rightarrow \frac{3x+z}{2z}+\frac{x+3z}{2x}$

$\Rightarrow \frac{1}{2}\left(\frac{3x^2+xz+xz+3z^2}{xz}\right)$

$\Rightarrow \frac{1}{2}\left(\frac{3x^2+3z^2+2xz}{xz}\right)$

Hence, this is the answer