+0

# Let x, y, and z be real numbers such that x+y+z, xy+xz+yz, and xyz are all positive.

0
197
15

Let $$x$$$$y$$, and $$z$$ be real numbers such that $$x+y+z$$$$xy+xz+yz$$, and $$xyz$$ are all positive.

Prove that $$x$$$$y$$, and $$z$$ all positive.

Mar 12, 2019
edited by Guest  Mar 12, 2019

#1
+234
+2

If they are all negative, then x + y + z would be negative, regardless of the values. Since xyz is positive, two of them aren't negative because a negative times a negative is a positive, and we've already ruled out the option of three. Since xy + xz + yz is positive, one isn't negative. Let's say x is the negative one. (It doesn't matter which one we choose because they appear at the same rate) If x is negative, then no matter the values of y and z, it will be negative because you're multiplying them by x. If you factor out the x, you get x( y + z) + yz. So whatever the value of y and z, it will be smaller than x(y + z). This rules out the one option. Sorry I can't explain the third part too well I don't think, It's a little hard to communicate the idea I had.

Hope this helps!

Mar 12, 2019
edited by LagTho  Mar 12, 2019
#2
0

It didn't help.

Guest Mar 12, 2019
#3
+234
+1

For example, if you have xy + xz -yz, no matter what you have for y and z, it will always be positive right? So reverse the signs, and you have xy - xz + yz, it will always be negative. So if x is negative, then xy + xz + yz will always be negative. So going through it, we've ruled out there being one negative variable, two, and three. So all of them must be positive.

LagTho  Mar 12, 2019
#5
+1

It still doesn't help.

Guest Mar 12, 2019
#8
+101751
0

It didn't help me either.  It is good that you had a go LagTho  but i think there are errors in your logic.

Melody  Mar 12, 2019
#9
+234
+1

Yeah, that's definitely possible, I've found that my methods are often unorthodox, and sometimes require some assuming. Sorry I couldn't be more help! Good job Guest for solving it!

LagTho  Mar 12, 2019
#10
+27912
0

I think Lag Tho's reasoning is correct, though his explanation might have been a little cumbersome!

Alan  Mar 12, 2019
#12
+101751
0

"Since xyz is positive, two of them aren't negative because a negative times a negative is a positive, and we've already ruled out the option of three."

If we had          a negative * a negative * a postitive then    xyz would be positive and that is what we want.

Melody  Mar 12, 2019
#15
+27912
+1

You’re right.  The explanation was even more confusing than I first thought!

Alan  Mar 12, 2019
edited by Alan  Mar 12, 2019
#4
+4

Let $$x$$, $$y$$, and $$z$$ be the roots of $$f(t)=at^3+bt^2+ct+d$$. We can divide by $$a$$ on both sides here because it doesn't affect the roots. This gives us
$$\frac{f(t)}{a}=t^3+\frac{b}{a}t^2+\frac{c}{a}t+\frac{d}{a}.$$
By Vieta's Formulas, $$x+y+z=-\frac{b}{a}$$, $$xy + xz + yz=\frac{c}{a}$$, and $$xyz=-\frac{d}{a}$$. Since $$x + y + z$$, $$xy + xz + yz$$, and $$xyz$$ are all positive, $$\frac{b}{a}$$ must be negative, $$\frac{c}{a}$$ must be positive, and $$\frac{d}{a}$$ must be negative.
For $$\frac{f(t)}{a}=0$$$$t$$ has to be positive. This is because if $$t$$ was negative, $$t^3$$ would be negative, $$\frac{b}{a}t^2$$ would be negative ($$t^2$$ is positive but $$\frac{b}{a}$$ is negative), $$\frac{c}{a}t$$ would be negative, and $$\frac{d}{a}$$ would be negative so $$\frac{f(t)}{a}$$ would have to be negative. If $$t$$ was $$0$$$$t^3$$ would be $$0$$, $$\frac{b}{a}t^2$$ would be $$0$$, $$\frac{c}{a}t$$ would be $$0$$, and $$\frac{d}{a}$$ would be negative so $$\frac{f(t)}{a}$$ would still be negative. Therefore, the only way for $$\frac{f(t)}{a}=0$$ is for $$t$$ to be positve. This means that all of $$f(t)$$'s roots (which are the same as $$\frac{f(t)}{a}$$'s roots) are positive so $$x, y,$$ and $$z$$ are all positive.

Mar 12, 2019
#6
0

Thank you!

Guest Mar 12, 2019
#7
0

Guest Mar 12, 2019
#14
+101751
0

Thanks guest. :)

Melody  Mar 12, 2019
#11
+2

Proof will be by contradiction, that is, we show that not just one of x, y, z can be negative, nor can just two of them be negative nor all three negative, leaving the only possibility that all three are positive.

If just one of x, y, z is negative then the product xyz would be negative, which is a contradiction.

If all three of x, y, z are negative, then the sum x + y + z would be negative, which again is a contradiction.

That leaves only the possibility that precisely two of the three are negative.

Suppose that, wlog, $$x = -u^{2}, \text{ and }y = -v^{2},\text{ with }z>0.$$

then, since x + y + z > 0, we have $$-u^{2}-v^{2}+z>0, \text{ so }z > u^{2}+v^{2}\dots(1).$$

From the second condition,

$$\displaystyle u^{2}v^{2}-u^{2}z-v^{2}z>0,\text{ so }\:u^{2}v^{2}>z(u^{2}+v^{2})\text{ and }\:\frac{u^{2}v^{2}}{u^{2}+v^{2}}>z\dots(2).$$

From (1) and (2),

$$\displaystyle \frac{u^{2}v^{2}}{u^{2}+v^{2}}>z>u^{2}+v^{2}, \text{ so }$$

$$\displaystyle u^{2}v^{2}>(u^{2}+v^{2})^{2}=u^{4}+v^{4}+2u^{2}v^{2}$$

again a contradiction, forcing the conclusion that all three are positive.

Tiggsy

Mar 12, 2019
#13
+101751
0

Thanks Tiggsy.

I do wish you would get yourself an account.

I have often want to look back at your contributions and I can't.

If you had posted under username it would be easy for me to do.

Melody  Mar 12, 2019