2
$\begingroup$

The following question is given in V.K Rohatgi problem 7.2.4.

Let $x_1,x_2,...,x_n$ be real numbers, and let $x_{(n)} = \max(x_1,x_2,...,x_n) $ for $n\geq 2$, and $x_{(1)} = \min(x_1,x_2,...,x_n)$. Show that for any set of real numbers $a_1,a_2,...,a_n$ such that $\sum_{i=1}^n a_i = 0$, the following inequality holds: $$\left|\sum_{i=1}^n a_i.x_i\right| \leq \frac{1}{2}(x_{(n)} - x_{(1)})\sum_{i=1}^n |a_i|$$

$\endgroup$
  • $\begingroup$ you meant $\sum_{i=1}^{n}$ likely $\endgroup$ – phdmba7of12 Sep 5 '17 at 14:16
  • 1
    $\begingroup$ yeah, my bad i'll correct it asap $\endgroup$ – AshishSinha5 Sep 5 '17 at 14:18
2
$\begingroup$

We assume without loss of generality that all the $a_i$ are non-zero.

Note that $a_i+|a_i|=0$ if $a_i< 0$ and $a_i+|a_i|=2a_i$ if $a_i> 0$.

Similarly, $a_i-|a_i|=0$ if $a_i> 0$ and $a_i-|a_i|=2a_i$ if $a_i< 0$.

Therefore, $\sum_i(|a_i|+a_i)=2\sum_{i, a_i> 0} a_i$ and $\sum_i(|a_i|-a_i)=-2\sum_{i, a_i< 0} a_i$. Using the hypothesis $\sum_i a_i = 0$, we get $$\sum_i|a_i| = 2\sum_{i, a_i> 0} a_i=-2\sum_{i, a_i< 0} a_i$$


If $a_i<0$, $a_ix_{(n)}\leq a_ix_i\leq a_ix_{(1)}$, hence $$x_{(n)}\sum_{i, a_i< 0} a_i \leq \sum_{i, a_i< 0}a_ix_i \leq x_{(1)}\sum_{i, a_i< 0} a_i $$

Similarly, $$x_{(1)}\sum_{i, a_i> 0} a_i \leq \sum_{i, a_i> 0}a_ix_i \leq x_{(n)}\sum_{i, a_i> 0} a_i $$

Summing these last two inequalities, $$x_{(1)}\sum_{i, a_i> 0} a_i + x_{(n)}\sum_{i, a_i< 0} a_i\leq \sum_{i}a_ix_i \leq x_{(1)}\sum_{i, a_i< 0} a_i + x_{(n)}\sum_{i, a_i> 0} a_i$$

But $ \displaystyle x_{(1)}\sum_{i, a_i< 0} a_i + x_{(n)}\sum_{i, a_i> 0} a_i = x_{(1)}\left(-\frac{1}2 \sum_i |a_i| \right)+x_{(n)}\left(\frac{1}2 \sum_i |a_i| \right) = \frac {x_{(n)}-x_{(1)}}2 \sum_i |a_i|$

Similarly, $\displaystyle x_{(1)}\sum_{i, a_i> 0} a_i + x_{(n)}\sum_{i, a_i< 0} a_i = -\frac {x_{(n)}-x_{(1)}}2 \sum_i |a_i|$

Hence $$-\frac {x_{(n)}-x_{(1)}}2 \sum_i |a_i|\leq \sum_{i}a_ix_i \leq \frac {x_{(n)}-x_{(1)}}2 \sum_i |a_i|$$

Lastly, $$\left|\sum_{i} a_ix_i\right| \leq \frac{x_{(n)} - x_{(1)}}{2}\sum_{i} |a_i|$$

$\endgroup$
  • $\begingroup$ @Glen_b what is bothering you ? $\endgroup$ – Gabriel Romon Sep 6 '17 at 6:05
  • $\begingroup$ @Glen_b oh yeah, of course, my bad ! $\endgroup$ – Gabriel Romon Sep 6 '17 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.