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Impossible inequality

  1. Jun 15, 2008 #1
    To the impossible ineqaulity of another thread let me add two new ones
    1)1/x +1/y +1/z >= 2/x+y + 2/y+z +2/z+x for x,y,z positive real Nos
    2) (x+y)'/x+y+2z + (y+z)'/y+z+2x + (z+x)'/z+x+2y>= (sqrt(x)+ sqrt(y)+sqrt(z))'/3
    where a' means a to the square and sqrt(x) means the sqaure root of x
  2. jcsd
  3. Jun 16, 2008 #2


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    First clear up your notation: use parentheses and "^2" is standard for "square".

    1)1/x+ 1/y+ 1/z>= 2/(x+y)+ 2/(y+z)+ 2/(z+ x)

    2) (x+y)^2/(x+ y+ 2z)+ (y+z)^2/(y+ z+ 2x)+ (z+x)^2/(z+ x+ 2y)>= (sqrt(x)+ sqrt(y)+ sqrt(z))^2/3.

    Now, are you asserting that these are identities for all positive real numbers or are they to be solved for specific x, y, z?
  4. Jun 16, 2008 #3
    Sorry for the icovinience ,for all x,y,z belonging to real Nos that is the +VE ones
  5. Jun 16, 2008 #4
  6. Jun 16, 2008 #5
    I'll make it pretty.

    [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{x+z} [/tex]

    [tex]\forall x, y, z \in \mathbb{R}_+[/tex]


    [tex]\frac{(x+y)^2}{x+y+2z}+\frac{(y+z)^2}{2x+y+z}+\frac{(x+z)^2}{x+2y+z}\geq (\sqrt{x} +\sqrt{y} +\sqrt{z})^{\frac{2}{3}}[/tex]

    I think again [tex]\forall x, y, z \in \mathbb{R}_+[/tex]
  7. Jun 16, 2008 #6
    (x+y)^2/x+y+2z +(y+z)^2/y+z+2x +(z+x)^2/z+x+2y>= (sqrt(x) +sqrt(y)+sqrt(z))^2/3
    it is sqrt(x) +sqrt(y)+sqrt(z) all to the square and all that divided by 3
  8. Jun 17, 2008 #7
    [tex]\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{x+z} [/tex]

    [tex]\forall x, y, z \in \mathbb{R}_+[/tex]


    [tex]\forall x, y, z \in \mathbb{R}_+[/tex]
    [tex]\frac{(x+y)^2}{x+y+2z}+\frac{(y+z)^2}{2x+y+z}+\frac{(x+z)^2}{x+2y+z}\geq \frac{(\sqrt{x} +\sqrt{y} +\sqrt{z})^{2}}{3}[/tex]
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