# Inequality related AM-GM

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.

• if you show your work it is more likely that you'll get an answer; you can still edit your question to improve it. Sep 20, 2015 at 17:42
• 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 Sep 20, 2015 at 18:08

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!