Strict subadditivity of the norm in uniformly convex Banach Spaces I would like to prove the following:

Let $(X,\|\cdot\|)$ be a uniformly convex Banach Space. Then the norm is strictly subadditive, i.e., $\forall x,y\in X\backslash\{0\}:\|x+y\|&lt;\|x\|+\|y\|\,$.

I started my proof by considering two elements satisfying $\|x+y\|=\|x\|+\|y\|$. By using all properties of a unifomly convex space it should follow that $x=y=0$, but I always failed. 

Can someone give me a hint how to proceed?
If $x$ is not a multiple of $y$, $\Vert y\Vert\ge\Vert x\Vert&gt;0$, and $\Vert x+y\Vert=\Vert x\Vert+\Vert y\Vert$, then
$$
\biggl\Vert  {x\over\Vert x\Vert}+ {y\over\Vert y\Vert} \biggr\Vert
\ge
\biggl\Vert  {x\over\Vert x\Vert}+ {y\over\Vert x\Vert} \biggr\Vert
-
\biggl\Vert  {y\over\Vert x\Vert}- {y\over\Vert y\Vert} \biggr\Vert
={{\Vert x\Vert +\Vert y\Vert}\over \Vert x\Vert}
-\Vert y\Vert\Bigl({1\over\Vert x\Vert}-{1\over \Vert y\Vert}\Bigr)=2.
$$

So $X$ is strictly convex if and only if for distinct norm one vectors $x$, $y$, $\Vert x+y\Vert&lt;2$.

$X$ is uniformly convex if and only if for each $\epsilon&gt;0$, there is a $\delta&gt;0$ so that for any $x$, $y$ in the closed unit ball of $X$ with $\Vert x-y\Vert\ge\epsilon$ one has $$\Bigl\Vert{ x+y\over 2}\Bigr\Vert\le 1-\delta.$$

It should be easy to see from the above that a uniformly convex normed space is strictly convex.
I think this property is not verified in uniformly convex Banach spaces in general, in fact it is not even true in the uniformly convex Banach space $(\mathbb{R},|.|)$:  $|1+2|=|1|+|2|$