The sharp constant is $\theta:=(2-\sqrt{2})/4<3/20$, so the answer is positive. An example when the minimum equals $\theta$ is provided by 4 vectors of length $1/4$ and distinct horizontal/vertical directions.

Assume that all these expressions exceed $\theta$. Draw the vectors from the origin, there are 4 angles between consecutive vectors (one of these angles may exceed $\pi$), which sum up to $2\pi$. Choose two opposite angles with sum not exceeding $\pi$, without loss of generality, these are the angles $2\alpha$ between $u$ and $v$ and $2\beta$ between $w$ and $x$, $\alpha+\beta\leqslant \pi/2$. We have $$\|u+v\|^2=\|u\|^2+\|v\|^2+2\|u\|\cdot \|v\|\cos 2\alpha=(\|u\|+\|v\|)^2-2\|u\|\cdot \|v\|(1-\cos 2\alpha)\\
\geqslant (\|u\|+\|v\|)^2-\frac{(\|u\|+\|v\|)^2}2(1-\cos2\alpha)=(\|u\|+\|v\|)^2\cos^2\alpha,$$
thus $\|u+v\|\geqslant (\|u\|+\|v\|)\cos \alpha$ and $$\theta<\|u\|+\|v\|-\|u+v\|\leqslant (\|u\|+\|v\|)(1-\cos\alpha),$$
analogously, $\theta<(\|w\|+\|x\|)(1-\cos \beta)$. This yields that $\alpha$, $\beta$ are strictly positive and $$
\frac1{1-\cos \alpha}+\frac1{1-\cos \beta}< \theta^{-1}(\|u\|+\|v\|)
+\theta^{-1}(\|w\|+\|x\|)=\theta^{-1}.$$
On the other hand, $f(x):=1/(1-\cos x)$ is a convex function on $(0,\pi/2)$
(indeed, $$f'(x)=\frac{-\sin x}{(1-\cos x)^2}=\frac{-2\sin \frac{x}2\cos \frac{x}2}{4\sin^4\frac{x}2}=-\frac12\frac{\cos \frac x2}{\sin^3\frac{x}2}$$
that increases on $(0,\pi/2)$ because $\sin x/2$ increases and $\cos x/2$ decreases.) Therefore, by Jensen inequality we have
$$
\frac1{1-\cos \alpha}+\frac1{1-\cos \beta}\geqslant \frac2{1-\cos \frac{\alpha+\beta}2}\geqslant \frac2{1-\cos\pi/4}=\theta^{-1},
$$
a contradiction.

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