# Impossible inequality

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

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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?

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

Inconvenience

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

(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

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

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

and

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