Euclidean norm second derivative I really need Your help.

I need to prove that Euclidean norm is strictly convex. I know that a function is strictly convex if $f ''(x)&gt;0$. Can I use it for Euclidean norm and how? $||x||''=\frac{||x||^2-x^2}{||x||^3}$

Thank You!
The function 
$$
\lVert \mbox{ }\rVert:\mathbb{R}^n\to \mathbb{R},\, f(x)=\lVert x\rVert
$$
is certainly convex. In fact, given $t\in [0,1]$ and $x,y\in \mathbb{R}^n$ we have:
$$
\lVert tx+(1-t)y\rVert=\lVert tx+(1-t)y\rVert\le \lVert tx\rVert+\lVert (1-t)y\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert.
$$
But is it STRICTLY CONVEX? i.e. if $t\in (0,1)$ and $x,y \in \mathbb{R}^n$, with $x\ne y$, then
$$
\lVert tx+(1-t)y\rVert&lt;t\Vert x\rVert+(1-t)\lVert y\rVert.
$$
The answer is NO. In fact, if $t\in (0,1)$, and $y=sx$, with $0&lt;s \ne 1$,  and $x\in \mathbb{R}^n$, we certainly have $x\ne y$. But
\begin{eqnarray}
\lVert tx+(1-t)y\rVert&amp;=&amp;\lVert tx+(1-t)sx\rVert=\lVert (t+(1-t)s)x\rVert=(t+(1-t)s)\lVert x\rVert\\
&amp;=&amp;t\lVert x\rVert+(1-t)\lVert sx\rVert=t\lVert x\rVert+(1-t)\lVert y\rVert
\end{eqnarray}
Hence the function $\lVert\mbox{ } \rVert$ is CONVEX but NOT STRICTLY CONVEX.
By the definition of convexity and the triangle inequality we have


  $$ ||ax+(1-a)y|| \leq |a| ||x||+ |1-a|||y||,\quad 0\leq a\leq1. $$