some properties of inner product space

Question

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

1. $|\langle u,v\rangle| \le || u|| + ||v||$

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

3. $|\langle u,v\rangle \le |\langle u,w \rangle | + |\langle w,v \rangle |$

4. $|| u +v|| \le ||u +w|| + ||w +v||$

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?

Answer 1

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 $u\neq0$.

Answer 2

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

$$\left | \langle x, y \rangle \right| \leq \frac 1 2 \left (\|x\|^2 + \|y\|^2 \right ).$$