tenniskidperson3 04.03.2013 19:57 Divide by $6^x$. We get $\left(\frac{1}{6}\right)^x+2\cdot\left(\frac{1}{3}\right)^x+3\cdot\left(\frac{1}{2}\right)^x<1$. But the left side is decreasing and equality happens at $x=2$, so the inequality is true for all $x>2$.