Someone seemed to have fun with valuations on number fields (see http://www.problem-solving.be/pen/viewtopic.php?t=216 ). Note that \[4a^{3}+2b^{3}+c^{3}-6abc = (\sqrt[3]{4}a+\sqrt[3]{2}b+c)(2 \sqrt[3]{2}a^{2}+\sqrt[3]{4}b^{2}+c^{2}-2ab-\sqrt[3]{2}bc-\sqrt[3]{4}ac ).\] Lets write simply $d$ for the second term on the right hand side. We see that $4a^{3}+2b^{3}+c^{3}-6abc =0$ has only the trivial solution $a=b=c=0$ by infinite descent $\mod 2$. Thus $|4a^{3}+2b^{3}+c^{3}-6abc | \geq 1 = \frac{|d|}{|d|}$ and division by $|d|$ gives us that $|\sqrt[3]{4}a+\sqrt[3]{2}b+c| \geq \frac 1{|d|}$. We are thus left to show that $|d| \leq 4a^{2}+3b^{2}+2c^{2}$, and I'm not in the mood to do this ($d=\| \sqrt[3]{4}\zeta_{3}^{2}a+\sqrt[3]{2}\zeta_{3}b+c \|^{2}>0$ is not hard to show). In general for any nonzero algebraic integer $z$, the norm $N(z)$ is a nonzero integer, thus $|N(z)| \geq 1$.

Peter wrote: Let $ a, b, c$ be integers, not all equal to $ 0$. Show that \[ \frac{1}{4a^{2}+3b^{2}+2c^{2}}\le\vert\sqrt[3]{4}a+\sqrt[3]{2}b+c\vert.\] The reference for this problem "Belarus 2001" is not correct. The correct reference is "CRUX(No. 7, Volume 25, 1999), Problem A240, Mohammed Aassila"

ideahitme wrote: ... The reference for this problem "Belarus 2001" is not correct. The correct reference is "CRUX(No. 7, Volume 25, 1999), Problem A240, Mohammed Aassila" Both references are incorrect. The problem appeared in the KöMaL magazine in December, 1996 (problem N.122.). See here. I was the author, but neither I can be sure that nobody ever constructed a similar exercise earlier...