Let $X$ be an inner product space. I want to show that $$ \langle u, u\rangle \langle v, w\rangle \le \langle u, v\rangle \langle u, w\rangle $$ for $u,v,w \in X$. I think this should be straightforward to show, but I couldn‘t come up with anything. I tried to first show
$$ \langle v, \langle u, u\rangle w - \langle u, w\rangle u\rangle \le 0 $$
but I was also stuck here after a while. I know the second argument here is orthogonal to $u$, but I couldn‘t make use of it. Any help is appreciated!
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Another way to see why this is not true: Working in $\mathbb{R}^3$ (this isn't that big a restriction, just take the span of general position $u,v,w$ and impose an orientation) so we have nice triple products, we find \begin{align*} \mathbf{w}\cdot (\mathbf{u}\times(\mathbf{v}\times \mathbf{u}))&=\mathbf{w}\cdot (\mathbf{v}(\mathbf{u}\cdot \mathbf{u})-\mathbf{u}(\mathbf{v}\cdot \mathbf{u}))\\ &=(\mathbf{w}\cdot \mathbf{v})(\mathbf{u}\cdot \mathbf{u})-(\mathbf{w}\cdot \mathbf{u})(\mathbf{v}\cdot \mathbf{u}) \end{align*} but of course the LHS can take any sign.
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The motivation behind this question was the following:
Let $S$ be a set of vectors with $\langle v,w \rangle \le 0$ for all $v,w\in S$. Hence the angle between any two vectors in this set is at least $\frac{\pi}{2}$. Consider the mapping $\varphi: v \mapsto \langle u,u\rangle v - \langle v,u\rangle u$ for some $u \in S$. I wanted to show that after applying $\phi$ to the elements of the set $S - \text{span}\{u\}$, the angles stay $\ge \frac{\pi}{2}$. Thanks to the answers posted I was able to see that this inequality holds in this case, only because of the additional condition $\langle v,w \rangle \le 0$ for all $v,w\in S$.
That is not true. Take $X=\mathbb{R}^2$, $u=(1,0)$ and $v=w=(0,1)$. Then $\langle u,u\rangle\langle v,w\rangle=1$, but $\langle u,v\rangle\langle u,w\rangle=0$.