# How to show an inequality in an inner product space?

Let $$V$$ be a real inner product space with inner product $$\langle\cdot\,,\cdot\rangle$$. For $$u,v,w\in V$$, how to show the following inequality $$\langle u,v\rangle \langle u,w\rangle\:\leq\: \frac{1}{2}\big(\langle v,w\rangle +\|v\|\|w\|\big)\,\|u\|^2\,?$$ I tried with Cauchy-Schwarz inequality but failed to prove the above.

It's clear that if at least one of $$\|u\|,\|v\|,\|w\|$$ is $$0$$ the inequality is true. Therefore we may assume without loss of generality that $$\|u\|,\|v\|,\|w\|$$ are all non-zero.
Define $$u'=\dfrac{u}{\|u\|}$$, $$v'=\dfrac{v}{\|v\|}$$ and $$w'=\dfrac{w}{\|w\|}$$. Dividing both sides by $$\|u\|^2\|v\|\|w\|$$ we get that the inequality is true for $$u,v,w$$ if and only if it is true for $$u',v',w'$$, therefore, we may assume that $$\|u\|=\|v\|=\|w\|=1$$. Then we need to prove that $$2\langle u,v\rangle \langle u,w\rangle \leq 1+\langle v,w\rangle$$.
Now, if $$\lambda \in \mathbb{R}$$, we have\begin{align*}0 & \leq \|\lambda u+v+w\|^2 \\ & =\lambda ^2\|u\|^2+2\lambda (\langle u,v\rangle +\langle u,w\rangle )+(\|v\|^2+\|w\|^2+2\langle v,w\rangle ) \\ & =\lambda ^2+2(\langle u,v\rangle +\langle u,w\rangle )+2(1+\langle v,w\rangle ). \end{align*}This is a quadratic equation in $$\lambda$$ that has at most one real root, therefore, its discriminant $$\Delta$$ satisfies $$\Delta \leq 0$$. This implies that$$(\langle u,v\rangle +\langle u,w\rangle )^2\leq 2(1+\langle v,w\rangle ).$$Now, if $$a,b\in \mathbb{R}$$, we have that $$0\leq (a-b)^2=a^2-2ab+b^2$$, which implies that $$4ab\leq a^2+2ab+b^2=(a+b)^2$$. In particular, with $$a=\langle u,v\rangle$$ and $$b=\langle u,w\rangle$$, we get$$4\langle u,v\rangle \langle u,w\rangle \leq (\langle u,v\rangle +\langle u,w\rangle )^2\leq 2(1+\langle v,w\rangle ).$$Therefore,$$2\langle u,v\rangle \langle u,w\rangle \leq 1+\langle v,w\rangle$$as wanted.
W.l.o.g. you may assume $$\|u\|=1$$, and then view the LHS as $$\langle v,u\rangle\,\langle u,w\rangle \;=\; \big\langle v,u\langle u,w\rangle\big\rangle \;=\; \langle v,Pw\rangle$$ with $$P$$ denoting the orthogonal projector onto the subspace spanned by $$\,u$$. In fact, $$P$$ can be an arbitrary orthogonal projector for the following to work: The inequality to be proven is transformed into $$\big\langle v,(2P-\mathbb 1)w\big\rangle\;\leqslant\;\|v\|\|w\|\,,$$ and this follows from CBS combined with $$\|2P-\mathbb 1\|=1$$.
The operator $$2P-\mathbb 1$$ is the reflection by the hyperplane $$\operatorname{Im} P$$. It satisfies $$(2P-\mathbb 1)^2 =\mathbb 1$$, and it is both orthogonal and symmetric.