Question:- Let $a,b$ and $c$ be integers, not all equal to $0$. Show that

$$\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt[3]{4}a+\sqrt[3]{2}b+c|$$

This problem was proposed in a canadian journal. The presented solution is very hard to grasp. Is there a better way to prove this(may be using calculus)?

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  • $\begingroup$ why downvotes?? $\endgroup$
    – user6262
    Feb 23, 2021 at 7:31
  • $\begingroup$ Because the left-hand side is obviously no greater than $\frac 12$ and the right-hand side is at least $1$. $\endgroup$ Feb 23, 2021 at 7:39
  • $\begingroup$ I think they need $a,b,c>0$ to use AM-GM like that. But of course the solution doesn't really depend on it. $\endgroup$ Feb 23, 2021 at 7:40
  • $\begingroup$ @Robert Shore RHS is not at least $1$. If $a=1,b=-1,c=0$ then RHS is $0.327$ which is less than $1$ $\endgroup$
    – user6262
    Feb 23, 2021 at 7:52
  • $\begingroup$ @RobertShore How about $a=-6, b =2, c=7$, $|\sqrt[3]{4}a+\sqrt[3]{2}b+c| \approx 0.004564212$? $\endgroup$
    – River Li
    Feb 24, 2021 at 6:05

1 Answer 1


Essentially the same proof, without the error of assuming $a,b,c>0$ and motivating how we come up with the first note:

The algebraic integer $\alpha_0=\sqrt[3]4a+\sqrt[3]2b+c$ has conjugates $\alpha_1=\zeta^2\sqrt[3]4a+\zeta\sqrt[3]2b+c$ and $\alpha_2=\zeta^4\sqrt[3]4a+\zeta^2\sqrt[3]2b+c$, $\zeta$ a primitive cube root of unity.

Since $\alpha_0\neq 0$ for all $(a,b,c)\in\mathbb{Z}^3-\{(0,0,0)\}$, the complex conjugates $\alpha_1$ and $\alpha_2$ are nonzero, hence the real numbers $\alpha_0$ and its number-theoretic norm $$ N_{\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q}}(\alpha_0)=\alpha_0\alpha_1\alpha_2=4a^3+2b^3+c^3-6abc $$ have the same sign. Since $\alpha_0\neq 0$, we have $\lvert N_{\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q}}(\alpha_0)\rvert\geq 1$. So it remains to prove $$ \alpha_1\alpha_2\leq 4a^2+3b^2+2c^2. $$ Note that $\alpha_1\alpha_2=2\sqrt[3]2a^2+\dots$ is automatically positive, so the $-2\sqrt[3]2a^2-\dots\leq 4a^2+3b^2+2c^2$ part of the given proof is never required.


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