Disequazione irrazionale $\sqrt{3x+1}>\sqrt{x+5}$

Text

Solve the irrational inequality $\sqrt{3x+1}>\sqrt{x+5}.$

Solution

Existence conditions: $3x+1\ge 0\implies x\ge -\tfrac13$ and $x+5\ge 0\implies x\ge -5$; together they give $x\ge -\tfrac13$. Both sides are non-negative, so we square keeping the direction: $3x+1>x+5\implies 2x>4\implies x>2.$ Since $x>2$ satisfies the existence conditions: $\boxed{x>2.}$