some properties of inner product space Let $V$ be an inner product  space  and $u,v,w\in V$. Choose the correct option


*

*$|\langle u,v\rangle| \le || u|| + ||v||$

*$|\langle u,v\rangle| \le \frac{1}{2}(|| u||^2 + ||v||^2)$

*$|\langle u,v\rangle \le |\langle u,w \rangle | + |\langle w,v \rangle |$

*$|| u +v||  \le ||u +w|| + ||w +v||$
My attempt:
For  option $1)$ i take $u =(2,0)$ and $v= (3,0)$
now  $|\langle u,v\rangle| = \langle (2,0), (3,0) \rangle = 6$ and $||u|| =2, ||v|| =3$ that is option $1$ is  false
option $2$ is false take $u =(3,0)$ and $v =(0,3)$
option $3$ and $4 $ is true  by triangle inequality
Is it correct?
 A: Your answer to the second option is wrong. It is true because, by the Cauchy-Schwarz inequality, $\bigl\lvert\langle u,v\rangle\bigr\rvert\leqslant\lVert u\rVert.\lVert v\rVert$ and $\lVert u\rVert.\lVert v\rVert\leqslant\frac12\bigl(\lVert u\rVert^2+\lVert v\rVert^2\bigr)$ since\begin{align}\lVert u\rVert.\lVert v\rVert\leqslant\frac12\bigl(\lVert u\rVert^2+\lVert v\rVert^2\bigr)&\iff2\lVert u\rVert.\lVert v\rVert\leqslant\lVert u\rVert^2+\lVert v\rVert^2\\&\iff\lVert u\rVert^2+\lVert v\rVert^2-2\lVert u\rVert.\lVert v\rVert\geqslant0\\&\iff\bigl(\lVert u\rVert-\lVert v\rVert\bigr)^2\geqslant0.\end{align}
Actually, the third option is false. In $\mathbb{R}^3$, with its usual inner product, take $u=(1,0,0)$, $v=(1,1,0)$, and $w=(0,0,1)$.
And the fourth option is also false. Take $u=v=-w$, with $u\neq0$.
A: Option $(2)$ is true. By Cauchy-Schwarz inequality we have $$\left | \langle x,y \rangle \right | \leq \|x\| \|y\|.$$ 
Now by AM-GM inequality we have 
$$\|x\|\|y\| \leq \frac 1 2 \left (\|x\|^2 + \|y\|^2 \right ).$$
So for all $x,y \in V$ we have 
$$\left | \langle x, y \rangle \right| \leq \frac 1 2 \left (\|x\|^2 + \|y\|^2 \right ).$$
