Prove : $\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$ Problem :
Let $a,b\in \mathbb{Z}$ such that $a\neq 0,b\neq 0$ ; $\left | a \right |\leq 100,\left | b \right |\leq 100$. 

Prove that:
   $$\left | a\sqrt{2}+b\sqrt{3} \right |> \frac{1}{350}$$

Thanks :)
P/s : I have no ideas about this problem ! :(
 A: If $a$ and $b$ have the same sign ($0$ has the same sign as any integer for this purpose), then you have
$$\lvert a\sqrt{2} + b\sqrt{3}\rvert = \lvert a\rvert \sqrt{2} + \lvert b\rvert \sqrt{3} \geqslant \sqrt{2} > \frac{1}{350}.$$
So suppose $a$ and $b$ have opposite sign. Then
$$\lvert a\sqrt{2} + b\sqrt{3}\rvert = \frac{\lvert a\sqrt{2} + b\sqrt{3}\rvert\cdot\lvert a\sqrt{2} - b\sqrt{3}\rvert}{\lvert a\sqrt{2}-b\sqrt{3}\rvert} = \frac{\lvert 2a^2 - 3b^2\rvert}{\lvert a\rvert\sqrt{2} + \lvert b\rvert\sqrt{3}}.$$
Since $2a^2-3b^2$ is a nonzero integer, and $\lvert a\rvert, \lvert b\rvert \leqslant 100$, we have
$$\frac{\lvert 2a^2 - 3b^2\rvert}{\lvert a\rvert\sqrt{2} + \lvert b\rvert\sqrt{3}} \geqslant \frac{1}{100(\sqrt{2}+\sqrt{3})} >\frac{1}{315}.$$
A: Let $\lambda = a\sqrt{2} + b\sqrt{3}$ where $a$, $b$ not both zero and $|a|,|b| \le 100$. We have
$$\begin{array}{rrcl}
& ( \lambda - a\sqrt{2})^2 - 3b^2 &=& 0\\
\implies & \lambda^2 + 2a^2 - 3b^2 &=& \sqrt{8} a \lambda\\
\implies & \lambda^4 - (4a^2+6b^2)\lambda^2 + (2a^2-3b^2)^2 &=& 0\\
\iff & \lambda^2 (4a^2 + 6b^2) &=& (2a^2 - 3b^2)^2 + \lambda^4\\
\implies & |\lambda| &\ge& \frac{|2a^2 - 3b^2|}{\sqrt{4a^2+6b^2}}
\end{array}$$
Since $\sqrt{3/2}$ is irrational and $2a^2 - 3b^2$ is an integer,
$|2a^2 - 3b^2|$ is at least $1$ and hence
$$|\lambda| \ge \frac{1}{\sqrt{4\times 100^2+6\times 100^2}} = \frac{1}{100\sqrt{10}} > \frac{1}{350}$$
A: Hint: Since $|a^22 - b^23|\geq1$,
$$
|a\sqrt{2}+b\sqrt{3}|\geq{1\over |a\sqrt{2}-b\sqrt{3}|}\geq {1\over(\sqrt{2}+\sqrt{3})100}.
$$
