min(ax/(bx+1), ay/(by+1))​=min(ax,ay​)/(min(bx,​by)+1) Holds!

For any positive value of a, b, x and y, Numerically (through Matlab simulation) I can prove that the following equality holds.

min(ax/(bx+1), ay/(by+1))=min(ax,ay)/(min(bx,by)+1).

Can anyone please prove that the above equality holds through the mathematical derivation?

Thanks in advance >Kader

Answer 1

[Matt J]

 Let z=max(1/x,1/y)=1/min(x,y). Then,

    min(a*x/(b*x+1), a*y/(b*y+1))

        = min(a/(b+1/x), a/(b+1/y))

        = a/(b+z)

        = (a/z)/((b/z)+1)

        = a*min(x,y)/(b*min(x,y)+1)

        = min(a*x,a*y)/(min(b*x,b*y)+1)

[kader]

Thanks a lot. I really appreciate your effort and time.