Find Min ?????????????
find expression A = Min [x per unit (1-x)] + [y / unit (1-y)] with x + y = 1
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tìm MIn biểu thức A=[x/căn(1-x)]+[y/căn(1-y)] với x+y=1
-Anonymous
dễ c/m (ac+bd)^2 =<(a^2+b^2)(c^2+d^2) [bun nhiãcopki](*)
ta thay 1> x ,y> 0 thay x=1-y và y=1-x
đat a=can[x/cany] ,b=can[y/canx] , c=can[xcany],d=can[ycanx]
thay vào (*)
(x+y)^2=< A*[xcany + ycanx] (**)
lai ap dung(*) và chu ý x+y=1 =>xy =<1/4
(xcany+ycanx)=canx.canxy + cany.canxy=<can[(x+y)(2xy)]=<1/can2
tu (**) => A>= can2 vay Amin=can2 khi x=y=1/2
-Anonymous
ta thay 1> x ,y> 0 thay x=1-y và y=1-x
đat a=can[x/cany] ,b=can[y/canx] , c=can[xcany],d=can[ycanx]
thay vào (*)
(x+y)^2=< A*[xcany + ycanx] (**)
lai ap dung(*) và chu ý x+y=1 =>xy =<1/4
(xcany+ycanx)=canx.canxy + cany.canxy=<can[(x+y)(2xy)]=<1/can2
tu (**) => A>= can2 vay Amin=can2 khi x=y=1/2
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