0
$\begingroup$

If $a_i, b_i, p_i \ge 0$, we can show by Cauchy–Schwarz inequality that $$\left(\sum_i {a_i p_i}\right)\left(\sum_i {b_i p_i}\right) \ge \left(\sum_i {p_i\sqrt{a_i b_i}} \right)^2.$$ What happens when $p_i$ can take negative values, but $\sum_i {a_i p_i} \ge 0$, and $\sum_i {b_i p_i} \ge 0$. Can we still say that $$\left(\sum_i {a_i p_i}\right)\left(\sum_i {b_i p_i}\right) \ge \left(\sum_i {p_i\sqrt{a_i b_i}} \right)^2?$$ Can any other restrictions on $p_i$ ensure that the above result holds?

$\endgroup$
1
  • $\begingroup$ Kindly note that $a_i \ge 0$, and $b_i \ge 0$ $\forall i$, whereas $p_i \ge 0$ for some values of $i$ and $p_i <0$ for others. $\endgroup$
    – Math Lover
    Commented May 11, 2017 at 18:13

2 Answers 2

1
$\begingroup$

If $a_i,b_i>0$ and $p_i<0~\forall i$ then the following inequality will obviously hold: $$\left(\sum_ia_i|p_i|\right)\left(\sum_ib_i|p_i|\right)\geq\left(\sum_i|p_i|\sqrt{a_ib_i}\right)^2$$ $$\Rightarrow\left(\sum_ia_i(-p_i)\right)\left(\sum_ib_i(-p_i)\right)\geq\left(\sum_i(-p_i)\sqrt{a_ib_i}\right)^2~~~~~~~~~~~~~~~[since ~|p_i|=-p_i]$$ $$or,~\left(-\sum_ia_ip_i\right)\left(-\sum_ib_ip_i\right)\geq\left(-\sum_ip_i\sqrt{a_ib_i}\right)^2$$ $$or,~\left(\sum_ia_ip_i\right)\left(\sum_ib_ip_i\right)\geq\left(\sum_ip_i\sqrt{a_ib_i}\right)^2$$

$\endgroup$
4
  • $\begingroup$ this proof will not hold if $p_i>0$ for some $i$ $\endgroup$
    – QED
    Commented May 11, 2017 at 18:25
  • $\begingroup$ Yes, I agree that the above proof is valid only if $p_i \le 0$ for all values of $i$. But in the question I asked, $p_i$ can be positive, negative, or zero, but $\sum_i {a_i p_i} \ge 0$, and $\sum_i {b_i p_i} \ge 0$. Would last constraints help to establish the inequality? $\endgroup$
    – Math Lover
    Commented May 11, 2017 at 18:28
  • $\begingroup$ So the it is not necessary that $a_i,b_i\geq 0~\forall i$? The only constraint is that $\sum a_ip_i,\sum b_ip_i\geq0$? $\endgroup$
    – QED
    Commented May 11, 2017 at 18:31
  • $\begingroup$ $a_i, b_i \ge 0$ for all $i$. Additionally, $\sum_i {a_i p_i} \ge 0$, and $\sum_i {b_i p_i} \ge 0$. $\endgroup$
    – Math Lover
    Commented May 11, 2017 at 18:39
0
$\begingroup$

The inequality would hold whenever $p_ia_i>0,p_ib_i>0~\forall i$. Because then take $\tilde{x}=(\sqrt{a_1p_1},\sqrt{a_2p_2},\cdots,\sqrt{a_np_n})'$ and $$\tilde{y}=(\sqrt{b_1p_1},\sqrt{b_2p_2},\cdots,\sqrt{b_np_n})'$$, then we know $$\tilde{x}.\tilde{y}=||\tilde{x}||.||\tilde{y}||\cos{\theta}\leq||\tilde{x}||.||\tilde{y}||$$ where $\theta$ is the angle between $\tilde{x}$ and $\tilde{y}$

$\endgroup$
3
  • $\begingroup$ In the problem, $a_i, b_i \ge 0$, but $p_i$ can take on negative values. So $a_i p_i <0$, and $b_i p_i <0$. $\endgroup$
    – Math Lover
    Commented May 11, 2017 at 17:50
  • $\begingroup$ In that case its also quite easy...write $p_i$ as $-|p_i|$ $\endgroup$
    – QED
    Commented May 11, 2017 at 18:04
  • $\begingroup$ Could you please elaborate how we can use $-|p_i|$ to establish the inequality. $\endgroup$
    – Math Lover
    Commented May 11, 2017 at 18:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .