# Better way to prove $\frac{1}{4a^2+3b^2+2c^2}\leq |\sqrt{4}a+\sqrt{2}b+c|$

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

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

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)? • why downvotes?? Feb 23, 2021 at 7:31
• Because the left-hand side is obviously no greater than $\frac 12$ and the right-hand side is at least $1$. Feb 23, 2021 at 7:39
• 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. Feb 23, 2021 at 7:40
• @Robert Shore RHS is not at least $1$. If $a=1,b=-1,c=0$ then RHS is $0.327$ which is less than $1$ Feb 23, 2021 at 7:52
• @RobertShore How about $a=-6, b =2, c=7$, $|\sqrt{4}a+\sqrt{2}b+c| \approx 0.004564212$? Feb 24, 2021 at 6:05

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 $$\alpha_0=\sqrt4a+\sqrt2b+c$$ has conjugates $$\alpha_1=\zeta^2\sqrt4a+\zeta\sqrt2b+c$$ and $$\alpha_2=\zeta^4\sqrt4a+\zeta^2\sqrt2b+c$$, $$\zeta$$ 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 $$N_{\mathbb{Q}[\sqrt{2}]/\mathbb{Q}}(\alpha_0)=\alpha_0\alpha_1\alpha_2=4a^3+2b^3+c^3-6abc$$ have the same sign. Since $$\alpha_0\neq 0$$, we have $$\lvert N_{\mathbb{Q}[\sqrt{2}]/\mathbb{Q}}(\alpha_0)\rvert\geq 1$$. So it remains to prove $$\alpha_1\alpha_2\leq 4a^2+3b^2+2c^2.$$ Note that $$\alpha_1\alpha_2=2\sqrt2a^2+\dots$$ is automatically positive, so the $$-2\sqrt2a^2-\dots\leq 4a^2+3b^2+2c^2$$ part of the given proof is never required.