norm inequality, if $x^t e = 0$ with $e=(1,1,...,1)$ I'm trying to prove that if $x$  and $y$ are in $\mathbb R^n$ such that $\langle x,e\rangle=0 $ where $e=(1,1,...,1)$ then
$$|\langle x,y\rangle |\le||x||_1\frac{y_{\max}-y_{\min}}{2}\qquad \text{for all} \quad y$$ 
I think that is necessary to use norms inequalities and Cauchy–Schwarz  inequality or Hölder's inequality.
Can you help me, please?
 A: Proof is by induction. Suppose this holds for $n<N$. Since a rearrangement of 
$x_{k}^{\prime }s$ does not change the inequality we may suppose there
exists $m$ such that $\ x_{1},x_{2},...,x_{m}$ are positive and $%
x_{m+1},x_{m+1},...x_n$  are negative. (zero terms can be dropped). Now
let $u_{k}=x_{k},v_{k}=y_{k}$ for $1\leq k\leq m$ and $u_{m+1}=-\sum
_{k=m+1}^{n}x_{k}$. Let $v_{m+1}=\frac{\sum%
_{k=m+1}^{n}x_{k}y_{k}}{\sum_{k=m+1}^{n}x_{k}}$. We apply
induction hypothesis to the vectors $(u_{1},u_{2},...,u_{m+1})$ and $%
(v_{1},v_{2},...,v_{m+1})$.  Assume that at least two of the $x_{k}^{\prime
}s$ are negative so that $m+1<n$. We can conclude, by induction hypothesis,
that $$\left\vert \sum_{k=1}^{m+1}u_{k}v_{k}\right\vert \leq
(\sum_{k=1}^{m+1}\left\vert u_{k}\right\vert )\frac{v_{\max
}-v_{\min }}{2}$$. Note that $$v_{m+1}-v_{j}=\frac{\sum%
_{k=m+1}^{n}x_{k}(y_{k}-v_{j})}{\sum_{k=m+1}^{n}x_{k}}=\frac{%
\sum_{k=m+1}^{n}x_{k}(y_{k}-j_{j})}{\sum_{k=m+1}^{n}x_{k}}=%
\frac{\sum_{k=m+1}^{n}\left\vert x_{k}\right\vert (y_{k}-y_{j})}{%
\sum_{k=m+1}^{n}\left\vert x_{k}\right\vert }$$ $\leq y_{\max }-y_{\min }
$. It follows from this that $v_{\max }-v_{\min }\leq y_{\max }-y_{\min }$
so we have $$\left\vert \sum_{k=1}^{m+1}u_{k}v_{k}\right\vert \leq
(\sum_{k=1}^{m+1}\left\vert u_{k}\right\vert )\frac{y_{\max
}-y_{\min }}{2}$$. Also note that $$\sum_{k=1}^{m+1}\left\vert
u_{k}\right\vert
=\sum_{k=1}^{m+1}x_{k}-\sum_{k=m+1}^{n}x_{k}=\sum%
_{k=1}^{n}\left\vert x_{k}\right\vert $$ and $$\sum%
_{k=1}^{m+1}u_{k}v_{k}=\sum_{k=1}^{m}u_{k}v_{k}-\sum%
_{k=m+1}^{n}x_{k}\frac{\sum_{k=m+1}^{n}x_{k}y_{k}}{%
\sum_{k=m+1}^{n}x_{k}}=\sum_{k=1}^{n}x_{k}y_{k}$$. This
finishes the proof in this case. If $x_{1},x_{2},...,x_{n-1}$ are positive
and $x_{n}$ is negative then $$\sum_{k=1}^{n}x_{k}y_{k}=\sum
_{k=1}^{n-1}x_{k}y_{k}-(\sum_{k=1}^{n-1}x_{k})y_{n}=\sum
_{k=1}^{n-1}x_{k}(y_{k}-y_{n})$$ so $$\left\vert
\sum_{k=1}^{n}x_{k}y_{k}\right\vert \leq
\sum_{k=1}^{n-1}\left\vert x_{k}\right\vert (y_{\max }-y_{\min })$$
and $$\sum_{k=1}^{n-1}\left\vert x_{k}\right\vert =\frac{1}{2}%
\sum_{k=1}^{n}\left\vert x_{k}\right\vert $$. Hence the desired inequality holds in this case also. The inequality is obvious for $n=1$ so we have finished the proof.
