I have been struggling with a problem from my linear algebra class. It goes as follows:
Let $V$ be a vector space with $\langle\,\cdot \mid \cdot\,\rangle$ dot product defined, and let $u, v, w$ be three distinct vectors in $V$. Prove that $$ 2 \bigl| \langle u \mid v \rangle \bigr| \, \bigl| \langle u \mid w \rangle \bigr| \leq \|u\|^2 \bigl( \|v\| \, \|w\| + |\langle v \mid w \rangle \bigr|) $$
I have tried to find a vector with norm equal to $$ \|u\|^2 \bigl( \|v\| \, \|w\| + |\langle v \mid w \rangle \bigr|) - 2 \bigl| \langle u \mid v \rangle \bigr| \, \bigl| \langle u \mid w \rangle \bigr|, $$ but I haven't been able to do so. I have also got this equivalent expression: $$ 1 - 2 \cos(\alpha_1) \cos(\alpha_2) + \cos(\alpha_3) \geq 0, \ $$ where $\alpha_i$ are the angles between the vectors. Any ideas?
We may assume that $u,v$ and $w$ are unit vectors. The inequality in question then becomes $$ 2\big|\langle u,v\rangle \langle u,w\rangle\big|\le1+\big|\langle v,w\rangle\big|. $$ Let $V$ be the inner product space in question. If $V$ is finite-dimensional, embed it in a higher-dimensional inner product space, so that we may assume that $\dim V\ge3$. Then $\operatorname{span}\{u,v,w\}\subseteq W\subseteq V$ for some three-dimensional inner product space $W$. Let $\{u,x,y\}$ be an orthonormal basis of $W$. Then we may write $v=a_1u+b_1x+c_1y$ and $w=a_2u+b_2x+c_2y$ where $|a_i|^2+|b_i|^2+|c_i|^2=1$. By Cauchy-Schwarz inequality, $|a_1a_2|+|b_1b_2|+|c_1c_2|\le1$. Therefore \begin{align*} 2\big|\langle u,v\rangle \langle u,w\rangle\big| &=2|a_1a_2|\\ &\le1+|a_1a_2|-|b_1b_2|-|c_1c_2|\\ &\le1+|a_1a_2+b_1b_2+c_1c_2|\\ &=1+\big|\langle v,w\rangle\big|.\\ \end{align*}