TITLE: Lower bound on $|a+b \sqrt{2} + c \sqrt{3}|$
QUESTION [7 upvotes]: I have stumbled across this question:
Let $a,b,c$ be integers, not all $0$ such that $\max(|a|,|b|,|c|)<10^6$. Prove that $|a+b \sqrt{2} + c \sqrt{3}| > 10^{-21}$. 
Could anybody help by solving this? Elementary solution is preferred.

REPLY [8 votes]: Let $$\begin{cases}f_{1}=a+b\sqrt{2}+c\sqrt{3}\\
f_{2}=a-b\sqrt{2}+c\sqrt{3}\\
f_{3}=a-b\sqrt{2}-c\sqrt{3}\\
f_{4}=a+b\sqrt{2}-c\sqrt{3}\end{cases}$$
It is clear $$f_{1}f_{2}f_{3}f_{4}\in Z,a,b,c\in Z$$
since $a,b,c$ are integer,and not all 0 ,so $f_{k}\neq 0,k=1,2,3,4$.and Note
$$\max\{|a|,|b|,|c|\}<10^6\Longrightarrow |f_{k}|<10^7,k=1,2,3,4$$ so we have
$$|f_{1}f_{2}f_{3}f_{4}|\ge 1\Longrightarrow |f_{1}|\ge\dfrac{1}{|f_{2}f_{3}f_{4}|}>10^{-21}$$