# Absolute Value Inequalities

Absolute Value Inequalities are usually proved by the absolute value being greater than or equal to a certain value. The square of the value is equal to the square of its absolute value.

### Proof of Absolute Value Inequalities

Prove $|a| + |b| \ge |a+b|$.

\begin{aligned} \require{AMSsymbols} \require{color} |a| &\ge a \text{ and } |b| \ge b &\color{red} |3| = 3 \text{ or } |-3| > -3 \\ |a||b| &\ge ab \\ 2|a||b| &\ge 2ab \\ a^2 + 2|a||b| + b^2 &\ge a^2 + 2ab + b^2 \\ |a|^2 + 2|a||b| + |b|^2 &\ge (a + b)^2 &\color{red} a^2 = |a|^2 \\ |a|^2 + 2|a||b| + |b|^2 &\ge |a + b|^2 \\ (|a| + |b|)^2 &\ge |a + b|^2 \\ \therefore |a| + |b| &\ge |a + b| &\color{red} \text{square roots for both sides} \end{aligned}

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