Let $x_1 \ge x_2 \ge \dots \ge x_n \ge 0$ and $y_1 \ge y_2 \ge \dots \ge y_n \ge 0$ such that $$\forall k \in [\![1,n ]\!], \prod_{i=1}^k x_i \le \prod_{i=1}^k y_i$$ (with equality for $k=n$).
I want to show that $$\sum_{i=1}^n \ln ^2(x_i) \le \sum_{i=1}^n \ln^2 (y_i)$$
I don't know how to prove this result but I'm sure that it is somehow related to convexity methods : I think that we can prove it by using Weyl's majorant theorem but the proofs of this theorem are quite hard...
Adding the additional constraint that $x_n\geq 1$ and $y_n\geq 1$, it is true and follows from Corollary 1.10 in the book Trace Ideals and Their Applications by B. Simon, since $x\mapsto \ln(x)^2$ is continuous and increasing on $[1,\infty)$ and $x\mapsto \ln(e^x)^2 = |x|^2$ is convex. The proof indeed follows from a convexity argument.
Without the additional constraint, it is false, as noticed in comments : just take $x_1=\dots=x_{n-1}=y_1=\dots=y_n = 1$ and $x_{n}\in(0,1)$. Then the product inequality is verified as $x_n\leq 1$, but the sum inequality is not verified since $\ln(x_n)^2 > 0$.