Disequazione irrazionale $\sqrt{x+5}\ge x-1$

Text

Solve the irrational inequality $\sqrt{x+5}\ge x-1.$

Solution

Existence condition: $x+5\ge 0\implies x\ge -5$. It is of the form $\sqrt{A}\ge B$; we distinguish two cases.

• Case 1 ($B<0$, i.e. $x<1$): always true; with the existence condition $-5\le x<1$.
• Case 2 ($B\ge 0$, i.e. $x\ge 1$): $x+5\ge (x-1)^2=x^2-2x+1$, hence $x^2-3x-4\le 0$, i.e. $(x-4)(x+1)\le 0\implies -1\le x\le 4$. With $x\ge 1$: $1\le x\le 4$. Combining the two cases: $\boxed{-5\le x\le 4.}$