Does the following norm inequality hold?

Question

Let $\times$ denotes the cross product and $\cdot$ denotes the dot product. $\boldsymbol{e',e'',e''',r',r'',r'''}$ are vectors in $\mathbb{R}^3$. Does the following inequality hold? $$||\boldsymbol{e'}\times \boldsymbol{e''}\cdot (\boldsymbol{r'}-\boldsymbol{r''})||+||\boldsymbol{e''}\times \boldsymbol{e'''}\cdot (\boldsymbol{r''}-\boldsymbol{r'''})|| \geq ||\boldsymbol{e'}\times \boldsymbol{e'''}\cdot (\boldsymbol{r'}-\boldsymbol{r'''})||$$ If yes, how to prove it? Thanks!

Answer 1

No, it doesn't hold. Note that the norms in the inequality are absolute values, since the triple product is a real number. Choosing $\boldsymbol{e'}=(1,0,0),\boldsymbol{e''}=(0,1,0),\boldsymbol{e'''}=(0,0,1)$, we get that $\boldsymbol{e'}\times \boldsymbol{e''}=\boldsymbol{e'''}$,$\boldsymbol{e''}\times \boldsymbol{e'''}=\boldsymbol{e''}$ and $\boldsymbol{e'}\times \boldsymbol{e'''}=-\boldsymbol{e''}$. In addition, choose $\boldsymbol{r'}=(1,2,0)$, $\boldsymbol{r''}=(0,0,0)$ and $\boldsymbol{r'''}=(0,1,1)$. Then the proposed inequality becomes: $$|\boldsymbol{e'''}\cdot \boldsymbol{r'}|+|\boldsymbol{e'}\cdot \boldsymbol{r'''}|\geq |\boldsymbol{e''}\cdot (\boldsymbol{r'}-\boldsymbol{r'''})|$$ But $|\boldsymbol{e'''}\cdot \boldsymbol{r'}|=|\boldsymbol{e'}\cdot \boldsymbol{r'''}|=0$ and $|\boldsymbol{e''}\cdot (\boldsymbol{r'}-\boldsymbol{r'''})|=1$.