Is there any condition such that if $T$ is a $t$-norm, then $T(a,b)T(c,d)\geq T(ac,bd)$?

Question

Let $T : [0, 1]\times [0, 1] \rightarrow [0, 1]$. A $t$-norm is a function $T$ with properties:

1. $ T (x, 1) = x$

2. If $y\leq z$ then $T(x,y)\leq T(x,z)$

3. $T (x, y) = T (y, x) $

4. $T (x, T (y, z)) = T (T (x, y), z) $

Related to this question, Prove (or disprove) $T(a,b)T(c,d)\leq T(ac,bd)$? the inequality $T(a,b)T(c,d)\leq T(ac,bd)$ not always hold. It can be $T(a,b)T(c,d)\geq T(ac,bd)$.

My question:

Is there any condition such that this inequality holds? $$T(a,b)T(c,d)\geq T(ac,bd)$$