Show an inequality in an inner product space

Question

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.

Answer 1

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.