Cauchy–Schwarz Inequality for Negative Numbers 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?
If $a_i,b_i&gt;0$ and $p_i&lt;0~\forall i$ then the following inequality will obviously hold: $$\left(\sum_ia_i|p_i|\right)\left(\sum_ib_i|p_i|\right)\geq\left(\sum_i|p_i|\sqrt{a_ib_i}\right)^2$$
$$\Rightarrow\left(\sum_ia_i(-p_i)\right)\left(\sum_ib_i(-p_i)\right)\geq\left(\sum_i(-p_i)\sqrt{a_ib_i}\right)^2~~~~~~~~~~~~~~~[since ~|p_i|=-p_i]$$
$$or,~\left(-\sum_ia_ip_i\right)\left(-\sum_ib_ip_i\right)\geq\left(-\sum_ip_i\sqrt{a_ib_i}\right)^2$$
$$or,~\left(\sum_ia_ip_i\right)\left(\sum_ib_ip_i\right)\geq\left(\sum_ip_i\sqrt{a_ib_i}\right)^2$$
The inequality would hold whenever $p_ia_i&gt;0,p_ib_i&gt;0~\forall i$. 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 $\tilde{y}$