Show that for all vectors $u,v,w \in \mathbb R^3$: $$\left|u+v\right|+\left|u+w\right|+\left|w+v\right|\le\left|u\right|+\left|v\right|+\left|w\right|+\left|u+v+w\right|$$ I know that we should show that : $$\sqrt{\sum_{i=1}^{3}\left(u_{i}+v_{i}\right)^{2}}+\sqrt{\sum_{i=1}^{3}\left(u_{i}+w_{i}\right)^{2}}+\sqrt{\sum_{i=1}^{3}\left(w_{i}+v_{i}\right)^{2}}$$$$\le\sqrt{\sum_{i=1}^{3}\left(u_{i}\right)^{2}}+\sqrt{\sum_{i=1}^{3}\left(v_{i}\right)^{2}}+\sqrt{\sum_{i=1}^{3}\left(w_{i}\right)^{2}}+\sqrt{\sum_{i=1}^{3}\left(u_{i}+v_{i}+w_{i}\right)^{2}}$$
Is that right?
But this way looks a little difficult and takes much time,does there any other way?
One can show by a straightforward (yet tedious) calculation that the following holds for $u,v,w\in\mathbb{R}^n:$
$$ |u|+|v|+|w|-|u+v|-|v+w|-|w+u|+|u+v+w|=\frac{\sum_{\text{cyc}}(|v|+|w|-|v+w|)\cdot(|u|-|v+w|+|u+v+w|)}{|u|+|v|+|w|+|u+v+w|}$$
Evidently, the right-hand side is non-negative because of the triangle inequality. The desired result follows.
$\textbf{Note}$ The "cyc" subscript indicates that the sum is a "cyclic" sum; Suppose we have a cyclic sum as follows: $$S:=\sum_{\text{cyc}} \:f(a,b,c)$$ Then, $S=f(a,b,c)+ f(b,c,a)+ f(c,a,b)$. A cyclic sum cycles through all the arguments of the summand and then takes the sum of its values.