3- \frac{ \sqrt{18} }{4} x \leq \sqrt{8} x- \frac{1}{5} |*20
60-5\sqrt{9*2}x\leq20\sqrt{4*2}x-4
60-5*3\sqrt2x\leq20*2\sqrt2x-4
60-15\sqrt2x\leq40\sqrt2x-4
-15\sqrt2x-40\sqrt2x\leq-4-60
-55\sqrt2x\leq-64 |*(-1)
55\sqrt2x\geq64
x\geq\frac{64}{55\sqrt2}=\frac{64*\sqrt2}{55\sqrt2*\sqrt2}
x\geq \frac{64\sqrt2}{55*2}
x\geq \frac{32\sqrt2}{55}
x\in \langle \frac{32\sqrt2}{55};+\infty)