Absolute-value and fractional inequalities

Text

Solve the following inequalities:

• (a) $|x+2|+|x-2|\ge 0$
• (b) $\dfrac{2x-1}{5}-\dfrac{3-2x}{2}\le 4$
• (c) $\dfrac{x^2+7x+6}{x^2+x+1}\le 0$
• (d) $\dfrac{2}{1-2x}-\dfrac{1}{2}+x>-1$

Solution

(a) Sum of two absolute values: always $\ge 0$, so it holds for every $x$.

(b) Multiplying by $10$: $2(2x-1)-5(3-2x)\le 40 \implies 4x-2-15+10x\le 40 \implies 14x\le 57 \implies x\le \dfrac{57}{14}$.

(c) The denominator $x^2+x+1$ has $\Delta<0$, always positive; the numerator $x^2+7x+6=(x+1)(x+6)\le 0 \implies -6\le x\le -1$.

(d) $\dfrac{2}{1-2x}-\dfrac12+x+1>0 \implies \dfrac{-2x^2+\tfrac52}{1-2x}>0$, i.e. $\dfrac{5-4x^2}{2(1-2x)}>0$. Breakpoints $-\dfrac{\sqrt5}{2}$, $\dfrac12$ (excluded), $\dfrac{\sqrt5}{2}$. Sign study: $-\dfrac{\sqrt5}{2}<x<\dfrac12 \ \vee\ x>\dfrac{\sqrt5}{2}$.

$\boxed{\begin{aligned} &\text{(a) } \forall x\in\mathbb{R}; \quad \text{(b) } x\le \tfrac{57}{14}; \quad \text{(c) } -6\le x\le -1;\\ &\text{(d) } -\tfrac{\sqrt5}{2}<x<\tfrac12 \vee x>\tfrac{\sqrt5}{2}\end{aligned}}$