Solve for real $x$: $\sqrt{x^2-4x+3} \ge 2-x$ .

Question

Solve for real $x$: $\, \sqrt{x^2-4x+3} \ge 2-x$ .

First of all $x^2 – 4x + 3 \ge 0 \implies x \in (-\infty, 1] \cup [3, \infty)$ .

Clearly the interval $[3, \infty)$ satisfies the above inequality as the LHS gives non negative value whereas RHS gives negative value but I'm not sure about the range $(-\infty, 1]$ since in that case both LHS and RHS are giving non negative values .

Answer 1

When $x\leq 1$, one may conclude that \begin{align*} \sqrt{x^{2} - 4x + 3} < \sqrt{x^{2} - 4x + 4} = \sqrt{(2 - x)^{2}} = 2 - x \end{align*} whence we conclude the solution set is given by $S = [3,+\infty)$.