Can Impossible Inequalities Be Solved with Positive Real Numbers?

[LAVRANOS]

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

 

[HallsofIvy]

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

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?

 

[LAVRANOS]

Sorry for the icovinience ,for all x,y,z belonging to real Nos that is the +VE ones

 

[LAVRANOS]

Inconvenience

 

[Diffy]

I'll make it pretty.


\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{x+z}

\forall x, y, z \in \mathbb{R}_+

and

\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}}

I think again \forall x, y, z \in \mathbb{R}_+

 

[LAVRANOS]

(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
THANKS DIFFY

 

[LAVRANOS]

\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{x+z}

\forall x, y, z \in \mathbb{R}_+

and

\forall x, y, z \in \mathbb{R}_+
\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}