if X+y+z=1,

minimum value of xy(x+y)^2+yz(y+z)/^2+zx(z+x)^2

wherex,y,z are positive real numbers.

Question

if X+y+z=1, minimum value of xy(x+y)^2+yz(y+z)/^2+zx(z+x)^2 wherex,y,z are positive real numbers.

mycroft holmes · Answer

The expression can be made as small as we please. If e is a very small positive number let x=y=e, then z=1-2e. Then the expression is of the order of e. All we can say is that the lower bound is zero

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if X+y+z=1, minimum value of xy(x+y)^2+yz(y+z)/^2+zx(z+x)^2 wherex,y,z are positive real numbers.

if X+y+z=1,

minimum value of xy(x+y)^2+yz(y+z)/^2+zx(z+x)^2

wherex,y,z are positive real numbers.

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if X+y+z=1, minimum value of xy(x+y)^2+yz(y+z)/^2+zx(z+x)^2 wherex,y,z are positive real numbers.

if X+y+z=1, minimum value of xy(x+y)^2+yz(y+z)/^2+zx(z+x)^2 wherex,y,z are positive real numbers.

1 Answers

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if X+y+z=1,

minimum value of xy(x+y)^2+yz(y+z)/^2+zx(z+x)^2

wherex,y,z are positive real numbers.