# Prove that x/x+2y+z+ y/y+2z+x + z/z+2x+y>=3/4 if x,y,z>0?

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### 1 Answer

You should assume that `x+y+z=1` , hence, substituting 1 for `x+y+z` yields:

`x/(1+y)+ y/(1+z) + z/(1+x)gt=3/4 `

You should use Buniakovski's inequality such that:

`a/x + b/y + c/z gt= 1/(ax+by+cz)`

Using this inequality yields:

`x/(1+y)+ y/(1+z) + z/(1+x)gt=1/(x(1+y) + y(1+z) + z(1+x))`

`x/(1+y)+ y/(1+z) + z/(1+x)gt=1/(x+xy+y+yz+z+zx)`

Substituting 1 for x+y+z yields:

`x/(1+y)+ y/(1+z) + z/(1+x)gt=1/(1+xy+yz+zx)`

You should remember that `xy+yz+zx =lt ((x+y+z)^2)/3 = 1/3`

Substituting `1/3` for `xy+yz+zx` yields:

`x/(1+y)+ y/(1+z) + z/(1+x)gt=1/(1+1/3)`

`x/(1+y)+ y/(1+z) + z/(1+x)gt= 3/4`

**Hence, the las inequality proves that`x/(x+2y+z) + y/(y+2z+x) + z/(z+2x+y) gt= 3/4.` **

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