# Tricky inequality on norms of vectors (part 2)

This question is a sort of continuation of this other question of mine Tricky inequality on norms of vectors .

Let $v_0,v_1,w_0,w_1\in \mathbb{R}^2$ be four non-zero vectors such that:

• $v_0=(v_{0x},0)$, $v_1=(v_{1x},0)$ with $v_{0x}>0,v_{1x}>0$
• $w_0=(w_{0x},w_{0y})$, $w_1=(w_{1x},w_{1y})$ with $w_{0y}\ge 0,w_{1y}\ge 0$
• $\displaystyle{\frac{||v_1||}{||v_0||}> 1}$ and $\displaystyle{\frac{||v_1||}{||v_0||}> \max\{\frac{||w_1||}{||w_0||},\frac{||w_0||}{||w_1||}\}}$ and $\displaystyle{\frac{||v_1||}{||v_0||}> \max\{\frac{||v_1+w_1||}{||v_0+w_0||},\frac{||v_0+w_0||}{||v_1+w_1||}\}}$

Define $$v_t:=((1-t)v_{0x}+t*v_{1x},0),\quad w_t:=(1-t)w_0+t*w_1$$ for every $t\in [0,1]$.

Is it true that the following two inequalities

$$\frac{||v_{t_1}||}{||v_{t_0}||}\ge \max\{\frac{||w_{t_1}||}{||w_{t_0}||},\frac{||w_{t_0}||}{||w_{t_1}||}\} \quad \textit{and}\quad \frac{||v_{t_1}||}{||v_{t_0}||}\ge \max\{\frac{||v_{t_1}+w_{t_1}||}{||v_{t_0}+w_{t_0}||},\frac{||v_{t_0}+w_{t_0}||}{||w_{t_1}+w_{t_1}||}\}$$

are always satisfied for every $t_0,t_1\in [0,1]$, $t_0<t_1$?

I can't find any counterexample, but I either can't prove the claim

We construct a counterexample as follows. Put $w_0=(-1,1)$ and $w_1=(1,1)$. Then $$\|w_0\|=\|w_1\|=\sqrt{2},$$ so $$\frac{\|w_0\|}{\|w_1\|}=\frac{\|w_1\|}{\|w_0\|}=1.$$ But $$\|w_{1/2}\|=\| (1/2) w_0+(1/2) w_1\|=\|(0,1)\|=1.$$ Now take small positive numbers $\varepsilon$ and $\delta$ and put $v_{0x}=\varepsilon$ and $v_{1x}=\varepsilon(1+\delta)$. This satisfies two of the required inequalities. The third inequality transforms to

$$(1+\delta)^2>\frac{1+(1+\varepsilon)^2}{1+(1-\varepsilon(1+\delta))^2}$$

which is equivalent to

$$\varepsilon<\frac{\delta}{\sqrt{1+\delta^2}}.$$

On the other hand, for every $t_0<t_1\in [0,1]$,

$$\frac{||v_{t_1}||}{||v_{t_0}||}=\frac{1+t_1\delta}{1+t_0\delta}\le 1+\delta.$$

Thus for a counterexample for the first asked inequality with $t_1=1/2$ and $t_0=1$ it suffices to take any (positive) $\delta<\sqrt{2}-1$ and $\varepsilon<\frac{\delta}{\sqrt{1+\delta^2}}$.