# Proving inequality using absolute value

Let $$\mathbf{x}\in\mathbb{R}^n, \mathbf{y}\in\mathbb{R}^n$$. We are given the following assumptions: $$$$\tag{1} \text{argmax}_j\{ |x_j| - |x_i|\}\geq \lambda > 0, \quad \forall i$$$$ $$$$\tag{2} \Big| |y_j| - |x_j|\Big|\leq \frac{\lambda}{3} , \quad \forall j$$$$ Meaning that the largest absolute value element is of $$\mathbf{x}$$ is greater than all the others by some strictly positive value, and that all the elements of $$\mathbf{x}, \mathbf{y}$$ are distant no more that $$\lambda/3$$.

Question: Assume w.l.o.g that element $$k$$ is the largest in $$\mathbf{x}$$: $$k = \text{argmax}_j\ |x_j|$$.I would like to prove that $$$$\tag{3} |y_k| - |y_j|\geq \frac{\lambda}{3} , \quad \forall j.$$$$ I tried proving this using the triangle inequality, but didn't get exactly what I needed. What I have so far is: \begin{aligned} \lambda&\leq |x_k| - |x_j|=\Big||x_k| - |x_j|\Big|\leq \Big||x_k|-|y_k|\Big|+\Big||y_k| - |x_j|\Big| \leq \frac{\lambda}{3}+\Big||y_k| - |x_j|\Big| \\ &\leq \frac{\lambda}{3} +\Big||y_k| - |y_j|\Big| + \Big||y_j| - |x_j|\Big|\leq \frac{2\lambda}{3}+\Big||y_k| - |y_j|\Big| \end{aligned} where the first equality is since we know that $$|x_k|>|x_j|$$ and the other transitions use the triangle inequality and the problem assumptions. Overall we have $$\Big||y_k| -|y_j|\Big|\geq \frac{\lambda}{3}$$, however I'm still left with the "outer" absolute value wrapping both terms, so this doesn't fully prove what I want.

Can anyone assist in proving equation $$(3)$$?

It's easiest to be a little more careful in the computation to see this holds true without the outer absolute value. \begin{align} \lambda &< |x_k|-|x_j|\\ &= |x_k| - |y_k| + |y_k| - |x_j| \\ &\leq \big||x_k| - |y_k|\big| + |y_k|-|x_j|\\ &\leq \frac{\lambda}{3} + |y_k|-|x_j|\\ &= \frac{\lambda}{3} + |y_k|-|y_j|+|y_j|-|x_j|\\ &\leq \frac{\lambda}{3} + |y_k|-|y_j|+\big||y_j|-|x_j|\big|\\ &\leq \frac{2\lambda}{3} + |y_k| - |y_j| \end{align}