Lower bound on $|a+b \sqrt{2} + c \sqrt{3}|$

Let $0 < \epsilon \ll 1$ be any small and $M \gg 1$ be any large positive numbers. Let $\lambda$ be a number of the form $a + b\sqrt{2} + c\sqrt{3}$ where $a, b, c \in \mathbb{Z}$, not all zero such that

$$|a|, |b|, |c| < M \quad\text{ and }\quad |\lambda| = |a + b\sqrt{2}+c\sqrt{3}| < \epsilon $$

It is easy to see $\lambda$ is a root of the polynomial

$$\begin{align} & \left((x-a - b\sqrt{2})^2 - 3c^2\right)\left((x-a + b\sqrt{2})^2 - 3c^2\right)\\ = & \left((x-a)^2 +2b^2 - 3c^2\right)^2 - 8b^2(x-a)^2\\ = & (x-a)^4 - 2(2b^2+3c^2)(x-a)^2 +(2b^2-3c^2)^2\\ \end{align}$$ Apply MVT for $x$ between $0$ and $\lambda$, we can find a $\xi \in (0,1)$ such that

$$4((\xi \lambda - a)^2 - 2b^2 - 3c^2)(\xi \lambda - a)x = -\left[a^4 - 2(2b^2+3c^2)a^2+(2b^2-3c^2)^2\right]$$ It is clear we can bound the absolute value of LHS from above by

$$4\max( (M+\epsilon)^2, 5 M^2 ) (M+\epsilon)\epsilon = 20 M^2 (M+\epsilon)\epsilon $$

For $M = 10^6$ and $\epsilon = 10^{-20}$, this bound is about $0.2$. Since RHS is an integer, this forces it to be zero. i.e.

$$a^4 - 2(2b^2+3c^2)a^2+(2b^2-3c^2)^2 = 0 \quad\iff\quad (a^2 - 2b^2 -3c^2)^2 = 24b^2c^2 $$ Since $24$ is not a square, we have

$$\begin{align} a^2 - 2b^2 - 3c^2 = bc = 0 \implies & ( b = 0 \land a^2 = 3c^2 ) \lor ( c = 0 \land a^2 = 2b^2 )\\ \implies & a = b = c = 0 \end{align} $$ This contradicts with our assumption that $a, b, c$ are not all zero.

Conclusion: If $|a|, |b|, |c| \le 10^6$ and not all zero, then $| a + b\sqrt{2} + c\sqrt{3} | \ge 10^{-20} > 10^{-21}$.

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Random Notes

Please note that what derived above is a very poor estimate of the actual lower bound.
If I didn't make any mistake, for $|a|, |b|, |c| < M = 10^6$, not all zero, we have

$$| a + b\sqrt{2}+c\sqrt{3} | \ge |376852-24672\sqrt{2}-197431\sqrt{3}| \approx 1.39824525 \times 10^{-11}\\ \color{red}{\text{this is wrong, see update below}} $$

It seems in general, the greatest lower bound is of the order $O(M^{-2})$ instead of $O(M^{-3})$. In view of Roth's Theorem, this is what one should expect. Unfortunately, I have no idea how to prove this speculation rigorously!

Update 2019/09/30

As pointed out by O.S. Dawg, above lower bound is incorrect.
With $(a,b,c) = (96051,-616920,448258)$, one has

$$a+b\sqrt{2}+c\sqrt{3} \sim 3.352882344 \times 10^{-13}$$ beating the lower bound I have before. Please note that to get this number, I need to crank up the precision in the computation (the new number is computed with 100 digits accuracy).

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 $$|f_{1}f_{2}f_{3}f_{4}|\ge 1\Longrightarrow |f_{1}|\ge\dfrac{1}{|f_{2}f_{3}f_{4}|}>10^{-21}$$