Disequazione irrazionale $\sqrt{x-1}>x-3$

Text

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

Solution

Existence condition: $x-1\ge 0\implies x\ge 1$. An inequality of the form $\sqrt{A}>B$ splits into two cases.

• Case 1 ($B<0$, i.e. $x<3$): the root is $\ge 0$, so the inequality always holds. With the existence condition: $1\le x<3$.
• Case 2 ($B\ge 0$, i.e. $x\ge 3$): squaring, $x-1>(x-3)^2=x^2-6x+9$, hence $x^2-7x+10<0$, i.e. $(x-2)(x-5)<0\implies 2<x<5$. With $x\ge 3$: $3\le x<5$. Combining the two cases: $\boxed{1\le x<5.}$