Prove that $$0\le yz+zx+xy-2xyz\le{7\over27},$$ where $x,y$ and $z$ are non-negative real numbers satisfying $x+y+z=1$.

This problem is from IMO 1984 and I know that this is a well-known problem and there bunch of proofs in AoPS. So please do not close this question as a duplicate because I am looking for an alternative solution and I provide my thoughts.

But I came up withthe following idea: Consider symmetric polynomial $\sigma_1=x+y+z, \sigma_2=xy+yz+xz, \sigma_3=xyz$. One can show that the following inequlities are true: i) $\sigma_1^2\geq 3\sigma_2;$ ii) $\sigma_2^2\geq 3\sigma_1\sigma_2;$ iii) $\sigma_1\sigma_2\geq 9\sigma_3$;

Hence the desired inequality inequality can be written as $0\leq \sigma_2-2\sigma_3\leq \frac{7}{27}$.

  1. The LHS inequality follows easily from iii) since in our case $\sigma_1=1$ and iii) implies that $\sigma_2\geq 9\sigma_3$;
  1. But I have issues with the RHS inequality. $$yz+zx+xy-2xyz=(yz+zx+xy)(x+y+z)-2xyz=$$$$=y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+xyz.$$

AM-GM gives us that $xyz\leq \frac{1}{27}$ and we have $7$ terms in the last equality so I was wondering is it possible to prove it somehow from that point?

However, I can obtain a slightly weaker bound: $$=yz(y+z)+xz(x+z)+xy(x+y)+xyz\leq yz+xz+xy+xyz=$$$$=\sigma_2+\sigma_3\leq \frac{1}{3}+\frac{1}{27}=\frac{10}{27}.$$

  • $\begingroup$ You may use three degree Schur. Let $p = a + b + c = 1, q =a b + bc + ca, r =a bc$. We have $r \ge \frac{4pq - p^3}{9}$. $\endgroup$
    – River Li
    Apr 20, 2022 at 13:47
  • $\begingroup$ @RiverLi, what do you mean by 3 degree Schur? Can you clarify it please? $\endgroup$
    – RFZ
    Apr 20, 2022 at 18:56
  • $\begingroup$ 3 degree Schur: $a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)\ge 0$ (degree 3 polynomial in $a, b, c$). In pqr language, it is $p^3 - 4pq + 9r \ge 0$. $\endgroup$
    – River Li
    Apr 20, 2022 at 23:39

1 Answer 1


For the right inequality:

pqr method:

Let $p = a + b + c = 1, \, q = ab + bc + ca, \, r = abc$.

It suffices to prove that $q - 2r \le \frac{7}{27}$.

Using $r \ge \frac{4pq - p^3}{9} = \frac{4q - 1}{9}$ (three degree Schur), it suffices to prove that $$q - 2 \cdot \frac{4q - 1}{9} \le \frac{7}{27}$$ or $$q \le \frac13$$ which is true using $p^2 \ge 3q$.

We are done.


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