A symmetric positive definite, $\Vert x \Vert = \sqrt{x^tAx}$ norm? Equivalent to euclidean norm? Let $A \in \mathbb{R}^{n,n}$ be a symmetric positive definite matrix. We define $\Vert x \Vert_A = \sqrt{x^tAx}$. I want to prove that is a norm. Moreover I want to prove that it is equivalent to the euclidean norm.
I know that $\Vert x \Vert_A$ is always positive and it is 0 if and only only when x = 0, so the first property is satisfied. 
Moreover, it is also immediate  that $\Vert \alpha x \Vert_A $ = $\vert \alpha \vert \Vert x \Vert_A$.
However, I am struggling to prove that $$ \Vert v_1 + v_2 \Vert \leq \Vert v_1 \Vert + \Vert v_2 \Vert$$
I think I should like the proof of Schwarz inequality, but I do not see how to do that here. 
Supposing it is a norm, how do I prove the equivalence with the Euclidean metric? I know that I have to find two positive numbers m,M such that:
$$ m\Vert x \Vert \leq \Vert x \Vert_A \leq M\Vert x \Vert $$
Any tips?
 A: Since $A$ is real and symmetric, there is an orthogonal $U$ such that
$U^T A U = \Lambda = \operatorname{diag}(\lambda_1,..., \lambda_n)$.
Since $A >0$ then $\lambda_n > 0$. 
Let $\sqrt{\Lambda} = \operatorname{diag}(\sqrt{\lambda_1,}..., \sqrt{\lambda_n)}$, then we can write 
$\|x\|_A = \| \sqrt{\Lambda} U^Tx \|_2$ and since
$ \sqrt{\Lambda} U^T$ is invertible, all the properties follow from those of the
Euclidean norm.
Note that $\lambda_n \|x\|_2^2 \le \langle x, Ax \rangle \le \lambda_1 \|x\|_2^2$
(assuming the eigenvalues are ordered), so we see that
$\sqrt{\lambda_n} \|x\|_2 \le \|x\|_A  \le \sqrt{\lambda_1} \|x\|_2$
A: We want to prove that 
$$\|v_1+v_2 \|_A \le \|v_1\|_A+\|v_2\|_A$$
$$(v_1+v_2)^TA(v_1+v_2) \le v_1^TAv_1+v_2^TAv_2+2\sqrt{(v_1^TAv_1)(v_2^TAv_2)}$$
which is equivalent to 
$$(v_1^TAv_2 )^2\le (v_1^TAv_1)(v_2^TAv_2)$$
Let $A = LL^T$ be the cholesky factoriazation, the statement become
$$((L^Tv_1)^T(L^Tv_2))^2 \le  ((L^Tv_1)^T(L^Tv_1))((L^Tv_2)^T(L^Tv_2))$$
Let $u=L^Tv_1$ and $v=L^Tv_2$ and use Cauchy-Schwarz should help.
As for equivalence property, for $x$ non-zero,we have $$\lambda_{\min} \le \frac{x^TAx}{x^Tx}\le \lambda_{\max}$$
