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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