For any vector $w\in\mathbb{R}^n,$ let $\|w\|:=\|w\|\_\infty = \max\_i |w\_i|.$ Let $w\geq 0$ mean that $w$ is non-negative in each co-ordinate.


For $0<\alpha<\frac 14,$ I am interested in the quantity


$$f(\alpha):=\sup\_{u,v\in\mathbb{R}^n, u,v\geq 0, n\geq 1} \frac{\|(1-\alpha)u-\alpha v\|+\|(1-\alpha)v-\alpha u\|+2\alpha \|u+v\|}{\|u\|+\|v\|}~.$$


By the triangle inequality, we get $f(\alpha)\leq 1+2\alpha.$


Or in other words, for $\alpha = p(1-p)$ with $0<p<\frac 12,$ this means $$f(p(1-p)) \leq 1+2p-2p^2.$$


By choosing $n=2$ and $u = \begin{bmatrix} p(1-p) \\ p(1-p) \end{bmatrix}$ and $v = \begin{bmatrix} p^2 \\ (1-p)^2 \end{bmatrix},$ we get


$$f(p(1-p))\geq 1+p-2p^2.$$


I conjecture that $f(p(1-p)) \leq 1+p-2p^2.$ 


Any proof or counterexample?



Ratio of sums of $\infty$-norms of vectors
You can do no better than $1+\alpha$.


Lower bound: Take $u=(1,0,1)$ and $v=(0,1,1)$.


Upper bound, the nominator is bounded by $(1-\alpha)\|u\|+(1-\alpha)\|v\|+2\alpha(\|u\|+\|v\|)$.


