Prove:

**great_math** · January 9th 2009, 03:44 AM

Prove the following inequalities.

1) $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{z} + \frac{y}{x} + \frac{z}{y}$

2) $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$

**ADARSH** · January 9th 2009, 04:22 AM

What I have Done is WRONG as rightly pointed by Isomorphism

Originally Posted by great_math

Prove the following inequalities.
1) $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{z} + \frac{y}{x} + \frac{z}{y}$

2) $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$

1)

Here $[\frac{x}{z} + \frac{y}{x} + \frac{z}{y}]^2$

$ =\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} +2(1+1+1) $
Here first term ie;
$ \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} $
is greater than zero and so is second term
and
$ [\frac{x}{z} + \frac{y}{x} + \frac{z}{y}]^2 \ge[\frac{x}{z} + \frac{y}{x} + \frac{z}{y}] $
If still in troble feel free to ask

**Isomorphism** · January 9th 2009, 11:59 PM

Originally Posted by great_math

Prove the following inequalities.

1) $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{z} + \frac{y}{x} + \frac{z}{y}$

Call $\frac{x}{y} = a, \frac{y}{z} = b , \frac{z}{x} = c$ . Then observe that your inequality amounts to proving $a^2 + b^2 + c^2 \geq ab + bc + ca$ But $a^2 + b^2 + c^2 \geq ab + bc + ca \Leftrightarrow \frac12 ((a-b)^2 + (b-c)^2 + (c-a)^2) \geq 0$ , which is clearly true.

Originally Posted by great_math

2) $\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$

EDIT: This proof is wrong

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Multiply out throughout by $x^2y^2z^2$ and then the inequality reduces to $x^4z^2 + z^4y^2 + y^4 x^2 \geq x^3yz^2 + y^3zx^2 + z^3 x y^2$ , which is clearly true by Muirhead inequality since (4,2,0) majorises (3,2,1).
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Originally Posted by ADARSH

1)
Here $[\frac{x}{z} + \frac{y}{x} + \frac{z}{y}]^2$

$ =\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} +2(1+1+1) $

Thats wrong Adarsh, $\left(\frac{x}{z} + \frac{y}{x} + \frac{z}{y}\right)^2 =\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} +2\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right) $

**JaneBennet** · January 20th 2009, 04:44 PM

Originally Posted by Isomorphism

$x^4z^2 + z^4y^2 + y^4 x^2 \geq x^3yz^2 + y^3zx^2 + z^3 x y^2$ , which is clearly true by Muirhead inequality since (4,2,0) majorises (3,2,1).

Doesn’t Muirhead give $x^4y^2 + x^4z^2 + y^4x^2 + y^4z^2 + z^4x^2 + z^4y^2 \geq x^3y^2z + x^3z^2y + y^3x^2z + y^3z^2x + z^3x^2y + z^3y^2x$ instead?

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