It looks easy but I cannot find the solution.

Let $x,y,z$ be positive reals and satisfy $x^2+y^2+z^2=2$, prove that $$ \left(x^3+\sqrt{3}xyz\right)\left( y^3-\sqrt{3}xyz\right)\leqslant 1. $$

I invented this inequality about a month ago. Essentially it can be proved by brutal force ( homogenize the inequality and then massive expanding), by SOS ( the usual method of express the polynomial into sum of square), or the Lagrange Multiplier Method (very painful task).

I am looking for a smart solution where the use of computer is minimum.

  • $\begingroup$ if you show your work it is more likely that you'll get an answer; you can still edit your question to improve it. $\endgroup$
    – Giovanni
    Sep 20, 2015 at 17:42
  • $\begingroup$ I proved that for all positive $x,y,z$ satisfies the condition $x^2+y^2+z^2=2$, then the best $k$ for the inequality $\left(x^3+kxyz\right) \cdot \left(y^3-kxyz\right) \leqslant 1$ is $k=\sqrt{3}$ So you are looking at the hardest case. Also this is three variables inequality but the symmetry is only on 2 variables. It is unique among three variables inequality. Plus, equality achieve when $x=y=1$ and $z=0$. Ordering variable cannot work here $\endgroup$
    – HN_NH
    Sep 20, 2015 at 18:08

1 Answer 1


If $y^2-\sqrt3xz<0$ then our inequality is obviously true.

But for $y^2-\sqrt3xz\geq0$ by AM-GM we obtain: $$(x^3+\sqrt3xyz)(y^3-\sqrt3xyz)=xy(x^2+\sqrt3yz)(y^2-\sqrt3xz)\leq$$ $$\leq\frac{1}{27}(x^2+y^2+xy+\sqrt3yz-\sqrt3xz)^3=$$ $$=\frac{1}{216}(2x^2+2y^2+2xy+2\sqrt3yz-2\sqrt3xz)^3=$$ $$=\frac{1}{216}(6-x^2-y^2-3z^2+2xy+2\sqrt3yz-2\sqrt3xz)^3=$$ $$=\frac{1}{216}\left(6-(x-y+\sqrt3z)^2\right)^3\leq\frac{6^3}{216}=1.$$ Done!


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