Let $\cdot$ denotes the dot product and $||\boldsymbol{x}||$ denotes the $L^2-$norm of the vector $\boldsymbol{x}$. Suppose $\boldsymbol{a,b,c}$ are vectors in $\mathbb{R}^3$. Does the following inequality hold? $$\left\|\frac{a\cdot b}{||a||*||b||} \right\|+\left\|\frac{b\cdot c}{||b||*||c||} \right\| \geq \left\|\frac{a\cdot c}{||a||*||c||} \right\|$$ If yes, how to prove it? Thanks!
Edit: There are two additional conditions that I should add to the above setup
- $\boldsymbol{a,b,c}$ are not orthogonal to each other
- $\boldsymbol{a,b,c}$ are different vectors in $\mathbb{R}^3$
- $\boldsymbol{a,b,c}$ are not parallel to each other
No, take $a = c$ and $b$ orthogonal to $a$ (and hence also $c$) then the RHS is $1$ and the LHS is $0$.
Edit for the updated question
No the inequality still wont hold. Let $v \neq 0$ be any vector that is not parallel or perpendicular to $a$ and let $a^{\perp}$ be a vector orthogonal to $a$. Then for $\epsilon > 0$ define $b = a^{\perp} + \epsilon v$ and $c = a + \epsilon v$. Then the LHS and RHS are both continuous functions of $\epsilon$ and the vectors are different, not parallel and non perpendicular. We have $$ a \cdot b = \epsilon a \cdot v \\ a \cdot c = \|a\|^2 + \epsilon a \cdot v \\ b \cdot c = \epsilon (a^{\perp} \cdot v + v \cdot a + \epsilon v \cdot v) $$
Thinking about the limit as $\epsilon \to 0$ we see that the LHS goes to $0$ but the RHS does not. As we have continuity there must be some $\epsilon>0$ for which the inequality does not hold. You can use this construction with explicit vectors to find such an epsilon.