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.

Guest Mar 12, 2019

edited by
Guest
Mar 12, 2019

#1**+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!

LagTho Mar 12, 2019

#3**+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

#8**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**+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**0 **

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

Alan
Mar 12, 2019

#12**0 **

What about this statement Alan.

"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

#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.

Guest 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

Guest Mar 12, 2019