We have the theorem that:
If $u_k,v_k$ are positive real numbers for $k = 1,...,n$ and $\frac{1}{p} + \frac{1}{q} = 1$ with real numbers p and q, such that pq < 0 (i.e. p < 0 or q < 0), Holder's inequality becomes here:
$$(\sum_{k=1}^{n} u_k^p)^{\frac{1}{p}}(\sum_{k=1}^n v_k^q)^{\frac{1}{q}} \leq \sum_{k=1}^n u_kv_k,$$ with equality if and only if $au_k^p = b_k^q$ for $k = 1,2...,n$ where a and b are real nonnegative constants with at least one nonzero.
Proof Assume p < 0, and set $P = - \frac{p}{q}, Q = \frac{1}{q}$ then $\frac{1}{P} + \frac{1}{Q} = 1$ with $P > 0$ and $Q > 0.$ Therefore, by Holder's inequality, we have $$(\sum_{k=1}^{n} U_k^P)^{\frac{1}{P}}(\sum_{k=1}^n V_k^Q)^{\frac{1}{Q}} \geq \sum_{k=1}^n U_kV_k,$$ where $U_k > 0$ and $V_k > 0$ for $k = 1,...,n.$ The above inequality for $U_k = u_k^{-q}$ and $V_k = u_k^qv_k^q$ gives the result. The result follows similarly for $q < 0,$ and the proof of equality holds similarly for the proof of when we have equality in Holder's regular inequality.
I don't really understand how plugging in these values for inequality gives the result, and I also have no idea how to do a similar proof for equality. Just to demonstrate, the proof of equality for regular Holder's inequality is:
$$(\sum_{k=1}^n u_k^p)^{-1} \cdot u_k^p = (\sum_{k=1}^n v_k^q) \cdot v_k^q$$ for $ k = 1,2,...,n.$ Setting appropriate values of $a$ and $b$ gives that equality holds in Young's inequality when $|a|^p = |b|^q.$