# Statement(s) > [!NOTE] Theorem (Reverse Triangle Inequality for Absolute Value) > For any $a,b \in \mathbb{R}$ we have $|a-b|\geq \bigg | |a|-|b| \bigg |$where $|\cdot|$ denotes the [[Absolute Value Function|absolute value function]]. > [!NOTE] Theorem (Reverse Triangle Inequality for Euclidean Norm) > For all $x,y\in \mathbb{R}^n$, $\lVert x-y \rVert \geq \bigg \lvert \lVert x \rVert - \lVert y \rVert \bigg \rvert $where $\lVert \cdot \rVert$ denotes the [[Euclidean Norm|Euclidean norm]] and $|\cdot|$ the absolute value function. > [!NOTE] Theorem (Reverse Triangle for Metric Space) > Contents # Proof(s) ###### Proof of Absolute Value Satisfies Reverse Triangle Inequality Applying the [[Absolute Value Satisfies Triangle Inequality]] to $(a-b)+b$ gives $|a|\leq |a-b|+|b| \implies |a|-|b|\leq |a-b|$Applying it to $(b-a)+a$ gives $|b|\leq |a-b|+|a| \implies |b|-|a|\leq |a-b|$Applying the definition of the [[Absolute Value Function]], $|a-b| \geq \max(|a|-|b|, |b|-|a|)=||a|-|b||$. ###### Proof of Reserve Triangle for Euclidean Norm Since [[Euclidean Norm Satisfies Triangle Inequality]], we have $\lVert x \rVert = \lVert x-y + y \rVert \leq \lVert x-y \rVert + \lVert y \rVert \implies \lVert x \rVert - \lVert y \rVert \leq \lVert x-y \rVert $and similarly $\lVert y \rVert - \lVert x \rVert \leq \lVert x-y \rVert$. Hence $\lVert x-y \rVert \geq \max \{ \lVert x \rVert - \lVert y \rVert , \lVert y \rVert - \lVert x \rVert \} = \bigg| \lVert x \rVert - \lVert y \rVert \bigg | $