Irrational inequality

Problem

Solve the inequality: $\sqrt{x^2+1} > |x-1|$

Solution

Recognising the strategy. Both sides are $\ge 0$ (a square root and an absolute value): since $t\mapsto t^2$ is increasing on $[0,+\infty)$, we may square both sides preserving the direction, without splitting into cases. The radicand $x^2+1$ is always positive, so there are no existence conditions to impose. Squaring (recalling that $|x-1|^2 = (x-1)^2$): $x^2 + 1 > (x-1)^2 = x^2 - 2x + 1$ Simplifying $x^2$ and $1$: $0 > -2x \;\Longrightarrow\; 2x > 0 \;\Longrightarrow\; x > 0$ $\boxed{x > 0}$

Links

Topics: Unifying methods
Concepts: Irrational inequality · Monotonic function · Absolute value
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality