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 ! :(

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
$$\frac{\lvert 2a^2 - 3b^2\rvert}{\lvert a\rvert\sqrt{2} + \lvert b\rvert\sqrt{3}} \geqslant \frac{1}{100(\sqrt{2}+\sqrt{3})} >\frac{1}{315}.$$
