# Question on a vector inequality

Is it true that

\min\left( \begin{aligned} &\|\mathbf{u}\| + \|\mathbf{v}\| - \|\mathbf{u} + \mathbf{v}\|, \\ &\|\mathbf{u}\| + \|\mathbf{w}\| - \|\mathbf{u} + \mathbf{w}\|, \\ &\|\mathbf{u}\| + \|\mathbf{x}\| - \|\mathbf{u} + \mathbf{x}\|, \\ &\|\mathbf{v}\| + \|\mathbf{w}\| - \|\mathbf{v} + \mathbf{w}\|, \\ &\|\mathbf{v}\| + \|\mathbf{x}\| - \|\mathbf{v} + \mathbf{x}\|, \\ &\|\mathbf{w}\| + \|\mathbf{x}\| - \|\mathbf{w} + \mathbf{x}\| \end{aligned} \right) \leq \frac{3}{20}

for all $$\mathbf{u}, \mathbf{v}, \mathbf{w}, \mathbf{x} \in \mathbb{R}^2$$ with $$\|\mathbf{u}\| + \|\mathbf{v}\| + \|\mathbf{w}\| + \|\mathbf{x}\| = 1$$?

Any insights or assistance would be greatly appreciated. Thank you in advance!

• If you give the source of this expression it could possibly be better (otherwise the effort to give an answer may consist for 80% of going back to your starting point. This usually happens here). Commented Sep 5 at 11:45
• @PietroMajer It was revised. Thanks. Commented Sep 5 at 23:52
• I have done many computations and have not found any counterexamples. Commented Sep 5 at 23:59
• @PietroMajer Hello, what if describing the question in the original language seems more complex, and one needs to find a solution that is clear or concise? Thanks. Commented Sep 6 at 4:15
• I would assume that the maximal possible value of the $\min$ is $1/2-1/\sqrt8$, attained on $\pm \frac14 e_1$, $\pm \frac14 e_2$. Is there any counterexample to this? Commented Sep 6 at 13:53

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.