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My question was closed because I didn’t define what $pqr$ was and didn’t have the original problem, so I’ve reposted with the added information. I apologize if I wasn’t supposed to repost. A solution to the original problem would also be nice, although I would prefer if it continued on my approach of using $pqr$. I suspect that Schur’s inequality ($r \ge \frac{4pq - p^3}{9}$ using $pqr$ notation) will have to be used, but I wasn’t able to use it nicely.

$pqr$ notation means that $p = x + y + z$, $q = xy + xz + yz$, and $r = xyz$ where $x$, $y$, $z$ are positive real numbers. This is a pretty standard technique for proving symmetric inequalities.

Using pqr notation, assuming that $p^2 - 2q = 1$, prove that $$\frac{p^2}{2} - pr + r^2 \le \frac{65}{54}.$$

I was solving an inequality and it reduced down to this, but I wasn’t sure how to prove this.

Here was the original problem: Assuming that $x^2 + y^2 + z^2 = 1$, prove that $$(1 - xy)(1 - yz)(1 - zx) \ge \frac{8}{27}.$$

solasky
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Alternative proof using the pqr method.

From $p^2 - 2q = 1$ and $p^2 \ge 3q$, we have $$q = \frac{p^2 - 1}{2}, \quad 1 \le p \le \sqrt{3}. \tag{1}$$

Using $q^2 \ge 3pr$, it suffices to prove that $$\frac{p^2}{2} - pr + r\cdot \frac{q^2}{3p} \le \frac{65}{54}$$ or $$\frac{p^2}{2} - \frac{3p^2 - q^2}{3p}\cdot r \le \frac{65}{54}. \tag{2}$$

Using (1), we have $3p^2 - q^2 > 0$. From (2), using $r \ge \frac{4pq - p^3}{9}$ (degree three Schur inequality), it suffices to prove that $$\frac{p^2}{2} - \frac{3p^2 - q^2}{3p}\cdot \frac{4pq - p^3}{9} \le \frac{65}{54}$$ or (using $q = \frac{p^2 - 1}{2}$) $$\frac{1}{108}(3 - p^2)(p^4 - 13p^2 + 144) \ge 0$$ which is true.

We are done.

River Li
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