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

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kader
kader 2016-8-4
评论： kader 2016-8-4
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

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Matt J
Matt J 2016-8-4
编辑：Matt J 2016-8-4
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)

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