The $n=1$ (!) Case of the Ahlswede--Daykin Four Functions Theorem

Question

Assume that $\alpha_i,\beta_i,\gamma_i,\delta_i$ ($i=0,1$) are non-negative real numbers such that $$ \alpha_0\beta_0\le\gamma_0\delta_0\\ \alpha_0\beta_0\le\gamma_1\delta_0\\ \alpha_0\beta_1\le\gamma_1\delta_0\quad\text{(note the subscripts)}\\ \alpha_1\beta_1\le\gamma_1\delta_1. $$ We want to prove that $$ (\alpha_0+\alpha_1)(\beta_0+\beta_1)\le(\gamma_0+\gamma_1)(\delta_0+\delta_1). $$ The three proofs I have looked at all say it suffices to consider the case where $\alpha_0=\beta_0=\gamma_0=\delta_0=1$, but none of them explain why.

Can you?

Answer 1

Hint: You can reduce to the case you want by dividing the left-hand side by $\alpha_0\beta_0$ and the right-hand side by $\gamma_0\delta_0$ (notice all numbers are non-negative, you'll need to treat the cases with zeros separately).