The function is a norm on $X$

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Show that if $(X, \langle, \rangle)$ is a space with inner product then the function $x \mapsto ||x||:= \langle x, x \rangle^{\frac{1}{2}}$ is a norm on $X$.

I have tried the following:

From the fact that $(X, \langle, \rangle)$ is a space with inner product we have the following properties:

  • $\langle x, x \rangle \geq 0, \forall x \in X$ and $\langle x,x \rangle=0$ iff $x=0$.
  • $\langle x,y \rangle= \langle y,x \rangle$
  • $\langle \lambda x+ \mu y ,z \rangle= \lambda \langle x,z \rangle + \mu \langle y,z \rangle$

We want to show the following properties:

  • $||x|| \geq 0$
  • $||x||=0$ iff $x=0$
  • $||\lambda x||=|\lambda| ||x||$ for every $\lambda \in \mathbb{R}$ and $x \in X$
  • $||x+y|| \leq ||x||+||y||$

$\langle x,x \rangle \geq 0 \Rightarrow \langle x, x \rangle^{\frac{1}{2}} \geq 0$ so the first property is satisfied.

$\langle x,x \rangle^{\frac{1}{2}} = 0 \Rightarrow \langle x, x \rangle =0 \Rightarrow x=0$.

If $x=0$ then $\langle x,x \rangle=0 \Rightarrow \langle x,x \rangle^{\frac{1}{2}}=0$.

Thus the second property is satisfied.

$||\lambda x||=\langle \lambda x, \lambda x \rangle^{\frac{1}{2}}=(\lambda \langle x, \lambda x \rangle)^{\frac{1}{2}}=(\lambda^2 \langle x, x \rangle)^{\frac{1}{2}}=|\lambda| \langle x,x \rangle^{\frac{1}{2}}$.

Therefore, the third property is satisfied.

Is everything right so far? Or could I improve something?

$||x+y||= \langle x+y,x+y \rangle ^{\frac{1}{2}}= ( \langle x,x+y \rangle + \langle y,x+y \rangle)^{\frac{1}{2}}=( \langle x,x \rangle +2 \langle x,y \rangle+ \langle y,y \rangle)^{\frac{1}{2}}$

Is it right so far? How could we show that the latter is $\leq \langle x,x \rangle^{\frac{1}{2}}+ \langle y,y \rangle^{\frac{1}{2}}$ ?

EDIT:If we us Cauchy-Schwarz in order to show the last property then would we get the following?

$$||x+y||= \langle x+y, x+y \rangle^{\frac{1}{2}} \leq \langle x+y, x+y \rangle \cdot \langle x+y, x+y \rangle=||x||^2+ 2\langle x,y \rangle +||y||^2$$

Is it right so far? But if so, then wouldn't we have to show that $\langle x,y \rangle=0$, in this case?