Proposition 1: If $\displaystyle x+y+z=1$ then $\displaystyle xy+yz+xz<1/2$

Q1. Prove Proposition 1 is true

Q2. Prove Proposition 1 is false

There is a Q3 for when Q1 and Q2 have been settled.

RonL

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- Jun 24th 2007, 11:02 PM #1

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- Jun 24th 2007, 11:47 PM #2
If $\displaystyle x,y,z\in\mathbf{R}$ then

$\displaystyle 1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$

$\displaystyle \Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$

$\displaystyle \displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$. So the proposition is true.

If $\displaystyle x,y,z\in\mathbf{C}$ then let $\displaystyle x=i,y=-i,z=1$.

Then $\displaystyle \displaystyle xy+xz+yz=1>\frac{1}{2}$. So the proposition is false.

- Jun 25th 2007, 02:41 AM #3

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- Jun 25th 2007, 05:06 AM #4
For $\displaystyle x,y,z\in\mathbf{R}$ the inequality is not tight.

We have $\displaystyle x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$

$\displaystyle \Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$

$\displaystyle \Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$.

The equality stands for $\displaystyle x=y=z=\frac{1}{3}$.

- Jun 25th 2007, 05:22 AM #5

- Jun 25th 2007, 05:40 AM #6

- Jun 25th 2007, 07:37 AM #7

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- Jun 25th 2007, 06:34 PM #8

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$\displaystyle x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.

Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

- Jun 25th 2007, 06:45 PM #9

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Actually that is Cauchy-Swartz

- Jun 25th 2007, 08:06 PM #10

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Show us how.

I see how it follows from the Cauchy Scwartz inequality:

$\displaystyle

| \bold{x} \cdot \bold{y} |\le \| \bold{x} \|\ \| \bold{y} \|

$

Then putting $\displaystyle \bold{x}=(a,b,c)$ and $\displaystyle \bold{y}=(b,c,a)$, with $\displaystyle a, b, c \in \mathbb{R}$, we have:

$\displaystyle

ab + bc + ca \le |ab + bc + ca| \le \sqrt{a^2+b^2+c^2}\ \sqrt{b^2+c^2+a^2} = a^2+b^2+c^2

$

RonL

- Jun 25th 2007, 08:10 PM #11

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- Jun 25th 2007, 08:48 PM #12

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- Jun 25th 2007, 09:36 PM #13
The inequality $\displaystyle x^2+y^2+z^2\geq xy+yz+zx$ can be proved like this:

Multiplying with 2, the inequality is equivalent to

$\displaystyle 2x^2+2y^2+2z^2\geq 2xy+2yz+2zx\Leftrightarrow$

$\displaystyle \Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\geq 0\Leftrightarrow$

$\displaystyle \Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0$.

- Jun 26th 2007, 03:53 AM #14

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