find the min of this inequality I want to minimize $$\displaystyle\min_{x\in\Bbb R}\bigl\{\max\{|1-\lambda_{1}x|,|1-\lambda_{2}x|,\cdots,|1-\lambda_{n}x|\}\bigr\},$$ where $0<\lambda_1<\lambda_2<\cdots<\lambda_n.$
I think this result $x=\dfrac{2}{\lambda_1+\lambda_n}.$ I hope can someone can give me some good methods. Thank you.
 A: Denote $f_j(x)=|1-\lambda_jx|$ and $f(x)=\max_j f_j(x)$.
Each of these functions is made of two affine pieces. Drawing it is highly recommended. 
In particular, on $(-\infty,\frac{1}{\lambda_j}]$, it is equal to $1-\lambda_jx$ so its graph is the half-line starting at $(\frac{1}{\lambda_j},0)$ and passing by $(0,1)$. On $[\frac{1}{\lambda_j},+\infty)$, it is worth $\lambda_j x-1$ so its graph is the half-line starting at $(\frac{1}{\lambda_j},0)$ with slope $\lambda_j$.
On $(-\infty,0]$, $f(x)=f_n(x)$ and it is decreasing.
Find the intersection of $f_1$ and $f_n$ on $(0,+\infty)$. You get 
$$
x_0=\frac {2}{\lambda_1+\lambda_n}\qquad f_1(x_0)=f_n(x_0)=\frac{\lambda_n-\lambda_1}{\lambda_n+\lambda_1}.
$$
On $[0,x_0]$, we have $f(x)=f_1(x)$ and it is decreasing.
Finally, on $[x_0,+\infty)$, we find $f(x)=f_n(x)$ and it is increasing.
Therefore the minimum is attained at $x_0$ and we have
$$
\min_\mathbb{R} f=f\left( \frac{2}{\lambda_1+\lambda_n}\right)=\frac{\lambda_n-\lambda_1}{\lambda_n+\lambda_1}.
$$
Note: to prove the claims more carefully, you can do like this.
On $(-\infty,0]$, $0$ is on the left of the zeros of all $f_j$, so
$$
f_j(x)=1-\lambda_jx\leq 1-\lambda_nx=f_n(x)\quad\Rightarrow \quad f(x)=f_n(x)=1-\lambda_nx \quad\forall x\leq 0.
$$
On $[0,x_0]$, $f_n$ is below $f_1$ and we are on the left of the zeros of $f_j$ for $j\leq n-1$, so
$$
f_j(x)=1-\lambda_jx\leq 1-\lambda_1x\quad\Rightarrow\quad f(x)=f_1(x)=1-\lambda_1x \quad \forall x\in [0,x_0].
$$
Finally, on $[x_0,+\infty)$, every $f_j$ is below $f_n$ (I confess I'm waving hands a little here, but this can be made very explicit without difficulty by considering each interval $[1/\lambda_{j+1},1/\lambda_j]$). So
$$
f(x)=f_n(x)=\lambda_nx-1\quad\forall x\geq x_0.
$$
So $f$ is continuous, decreasing on $(-\infty,x_0]$, increasing on $[x_0,+\infty)$. Hence $\min f=f(x_0)$.
A: this problem from this:we have $||x_{k}-x^{*}||_{A}\le \displaystyle\max_{1\le i\le n}{|1-\alpha\lambda_{i}}|\cdot||x_{k-1}-x^{*}||_{A}
\le\displaystyle\max_{\lambda_{n}\le \lambda\le\lambda_{1}}{|1-\alpha\lambda|}\cdot||x_{k-1}-x^{*}||_{A}$,and $\lambda_{1}>\lambda_{2}>\cdots>\lambda_{n}$
We made the first on the right formula to minimize,we easy
$a=\dfrac{2}{\lambda_{1}+\lambda_{n}}$
