Inspired by Jake Lai

Using trapezoids, we see that $\ln2=\int_{1}^{2}\frac{dt}{t}<\frac{3}{4}$ and $\ln3<\frac{4}{3}$ . Thus $\ln3<(\ln2)^{-1}$ and $\ln\ln3<-\ln\ln2$ . Since $\ln\ln3>0$ , we can square both sides to find that $(\ln\ln3)^2<(\ln\ln2)^2$

This solution was based on the assumption that $\ln$ refers to $\log$ base 10.

0 < ln(2)<ln(3) < 1 Let ln(2)=a

ln(3)=b

Then ln(a)>ln(b) ( as ln(x) is decreasing function b/w 0 and 1)

Thus ln(a)^2>ln(b)^2