Inequality Problem (A.M.-G.M)

[Kagami_Yoru01]

If x > y > 0, find the minimum of x + 8/(y(x-y)).
I have been trying to solve it for a long time now. And still can't figure it out; the problem is I can't make a new term which can be cancelled and give me a constant value. Can anybody give me a hint or a way to solve it?

** This is about A.M.-G.M. Inequalty problem

 

[BigBeachBanana]

If x > y > 0, find the minimum of x + 8/(y(x-y)).
I have been trying to solve it for a long time now. And still can't figure it out; the problem is I can't make a new term which can be cancelled and give me a constant value. Can anybody give me a hint or a way to solve it?

** This is about A.M.-G.M. Inequalty problem

Hint: If [imath]x>y>0[/imath], then [imath]y(x-y)\le?[/imath]

 

[BigBeachBanana]

Hint: If [imath]\cancel{x>y>0}[/imath], then [imath]\cancel{y(x-y)\le?}[/imath]

My previous hint only leads to a minimum bound, which requires additional work.
New Hint: Let [imath]z=x-y[/imath], then
[math]f=x+\frac{8}{y(x-y)}=x-y+y+\frac{8}{y(x-y)}=z+y+\frac{8}{yz}[/math]Apply AM-GM inequality...

 

[Kagami_Yoru01]

My previous hint only leads to a minimum bound, which requires additional work.
New Hint: Let [imath]z=x-y[/imath], then
[math]f=x+\frac{8}{y(x-y)}=x-y+y+\frac{8}{y(x-y)}=z+y+\frac{8}{yz}[/math]Apply AM-GM inequality...

@BigBeachBanana Wow! I didn't think about substitution method. Your reccommend helps me a lot. THANK YOU.