Problem 28

CaptainBlack · #1 $Old$ June 25th, 2007, 01:02 AM

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

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

red_dog · #2 $Old$ June 25th, 2007, 01:47 AM

If $x,y,z\in\mathbf{R}$ then
$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$
$\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$
$\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$ . So the proposition is true.
If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$ .
Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$ . So the proposition is false.

CaptainBlack · #3 $Old$ June 25th, 2007, 04:41 AM

Quote:

Originally Posted by red_dog $View Post$

If $x,y,z\in\mathbf{R}$ then
$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$
$\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$
$\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$ . So the proposition is true.
If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$ .
Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$ . So the proposition is false.

Q3. For $x,y,z \in \mathbb{R}$ is the inequality tight, if not can you find a tight version.

RonL

red_dog · #4 $Old$ June 25th, 2007, 07:06 AM

For $x,y,z\in\mathbf{R}$ the inequality is not tight.
We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$
$\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$
$\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$ .
The equality stands for $x=y=z=\frac{1}{3}$ .

janvdl · #5 $Old$ June 25th, 2007, 07:22 AM

Hehehe, looks like this guy knows what he's doing, eh CaptainBlack?

topsquark · #6 $Old$ June 25th, 2007, 07:40 AM

"tight"?

-Dan

CaptainBlack · #7 $Old$ June 25th, 2007, 09:37 AM

Quote:

Originally Posted by red_dog $View Post$

For $x,y,z\in\mathbf{R}$ the inequality is not tight.
We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$
$\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$
$\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$ .
The equality stands for $x=y=z=\frac{1}{3}$ .

You need to fill in some of the detail so others can follow this more easily.

RonL

mathisfun1 · #8 $Old$ June 25th, 2007, 08:34 PM

$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

ThePerfectHacker · #9 $Old$ June 25th, 2007, 08:45 PM

Quote:

Originally Posted by mathisfun1 $View Post$

$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

Actually that is Cauchy-Swartz

CaptainBlack · #10 $Old$ June 25th, 2007, 10:06 PM

Quote:

Originally Posted by mathisfun1 $View Post$

$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

Show us how.

I see how it follows from the Cauchy Scwartz inequality:

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

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

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

RonL

ThePerfectHacker · #11 $Old$ June 25th, 2007, 10:10 PM

Quote:

Originally Posted by CaptainBlank $View Post$

Show us how.

In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.

CaptainBlack · #12 $Old$ June 25th, 2007, 10:48 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.

May be, but he should still make it explicit.

Perhaps we should have a Wiki page on inequalities and their derivation/proof?

RonL

red_dog · #13 $Old$ June 25th, 2007, 11:36 PM

The inequality $x^2+y^2+z^2\geq xy+yz+zx$ can be proved like this:
Multiplying with 2, the inequality is equivalent to
$2x^2+2y^2+2z^2\geq 2xy+2yz+2zx\Leftrightarrow$
$\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\geq 0\Leftrightarrow$
$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0$ .

mathisfun1 · #14 $Old$ June 26th, 2007, 05:53 AM

According to AM-GM, $\frac{x^2+y^2}{2} \ge xy$ . Do the same for the other pairs of variables and add to get the desired inequality.

Credit must be given where credit is due -- I picked up this trick from the AoPS book Vol 2.