[tex]\sqrt{2} \cdot \sqrt{3} =\sqrt{2\cdot 3} =\sqrt{6} \\\\\\\dfrac{x+1}{\sqrt{3} } >\dfrac{x+1}{\sqrt{2} } ~~~~\mid \cdot \sqrt{2} \cdot \sqrt{3}\\\\\sqrt{2} \cdot (x+1)>\sqrt{3}\cdot (x+1)\\\\\sqrt{2}x+\sqrt{2}>\sqrt{3}x+\sqrt{3}\\\\\sqrt{2}x-\sqrt{3}x>\sqrt{3}-\sqrt{2} \\\\x\cdot (\sqrt{2}-\sqrt{3})>\sqrt{3}-\sqrt{2}\\\\-x\cdot (\sqrt{3}-\sqrt{2})>\sqrt{3}-\sqrt{2}~~\mid \div (\sqrt{3}-\sqrt{2})[/tex]
[tex]-x>1~~\mid \cdot (-1)\\\\x<-1~~~~\Rightarrow ~~x\in (-\infty ; -1)[/tex]
