Answer to Question #347867 in Analytic Geometry for Stella

Question

Prove that |a+b| - |a-b| ≤ 2|b|.

Answer 1

Consider paralelogram $ABCD$

$$\overrightarrow{AB}=\vec a, \overrightarrow{AD}=\vec b$$

Then

$$\overrightarrow{AO}=\dfrac{1}{2}(\vec{a}+\vec {b}), \overrightarrow{DO}=\dfrac{1}{2}(\vec{a}-\vec {b})$$

The sum of any two sides of a triangle is greater than or equal to the third side.

$$\triangle AOD:AD+OD\ge AO$$

Then

$$|\vec{b}|\ge |\dfrac{1}{2}(\vec{a}+\vec {b})|-|\dfrac{1}{2}(\vec{a}-\vec {b})|$$

$$|\vec{a}+\vec {b}|-|\vec{a}-\vec {b}|\le 2|\vec{b}|$$