Disequazione irrazionale $\sqrt{x-2}\ge\sqrt{4-x}$

Text

Solve the irrational inequality $\sqrt{x-2}\ge\sqrt{4-x}.$

Solution

Existence conditions: $x-2\ge 0\implies x\ge 2$ and $4-x\ge 0\implies x\le 4$; together $2\le x\le 4$. Both sides are non-negative, so we square: $x-2\ge 4-x\implies 2x\ge 6\implies x\ge 3.$ Intersecting with $2\le x\le 4$: $\boxed{3\le x\le 4.}$