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Let $x$, $y$, and $z$ be real numbers such that their symmetric sums $s_1$, $s_2$, and $s_3$ are all positive. In other words,

\begin{align*} s_1 &= x+y+z>0\\ s_2 &= xy+xz+yz>0\\ s_3 &= xyz>0 \end{align*}

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


I first tried proof by contradiction, assuming that at least one of $x$, $y$, and $z$ are negative. I also assumed that $x<y<z$, because the symmetric sums are (as the name implies) symmetric. If $y<z<x$, the proof should still be valid. Unfortunately, I couldn't get any farther than this.

I then tried finding $s_1$, $s_2$, and $s_3$, first by trying to multiply them and find relationships (I ended up with $s_1s_2s_3=s_3^2+s_3((x+y)(x+z)(y+z)-2s_3)$, which was unhelpful).

After that I tried brute force substitution, which didn't work either.

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When you deal with symmetric sums, always think of Vieta! Because $s_1,s_2,s_3$ are the symmetric sums of the respective order of $x,y,z$ if and only if $x,y,z$ are the roots of the polynomial $p(t)=t^3-s_1t^2+s_2t-s_3$.

Now it suffices to observe that $p(t) < 0$ whenever $t\leq 0$, which is evident.

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Let $x\geq y\geq z$.

Thus, since $x+y+z>0$, we obtain $x>0$.

Hence, since $xyz>0$, we see that $yz>0$.

Now, if $y>0$ and $z>0$ we are done.

Let $y<0$and $z<0$.

Now, we can replace $y\rightarrow-y$ and $z\rightarrow-z$.

Thus, $$x-y-z>0$$ and $$yz-xy-xz>0,$$ where $y$ and $z$ they are positives, which is impossible because $$yz>x(y+z)>(y+z)^2,$$ which gives $$y^2+yz+z^2<0,$$ which is absurd.

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  • $\begingroup$ @King Tut I added something. See now. $\endgroup$ – Michael Rozenberg Mar 2 '18 at 15:13
  • $\begingroup$ Because new $x$, $y$ and $z$ are positives and $x>y+z$. See please better my post. $\endgroup$ – Michael Rozenberg Mar 2 '18 at 15:18
  • $\begingroup$ Oh sorry didn't notice that you transformed the variable! Nice :) $\endgroup$ – King Tut Mar 2 '18 at 15:20

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