# $A_1, A_2\in [AB]$ s.t. $AA_1 < AA_2 < 1$

Let $ABCD$ a square of length $2$.

If we consider $A_1, A_2\in [AB]$ s.t. $AA_1 < AA_2 < 1$ and

$B_1, B_2\in [BC]$ s.t. $BB_1 < BB_2< 1$ and $C_1, C_2\in [CD]$

$D_1, D_2\in [DA]$ with the same properties then show that

$A_1B_1+B_1C_1+C_1D_1 +D_1A_1 > A_2B_2+B_2C_2+C_2D_2+D_2A_2$.

Lemma. Let $ABCD$ be a square with edge of length $2$. Let $P\in[AB]$ with $AP$<1, $X,X'\in[BC]$ with $BX<BX'<1$ and $Q\in[CD]$ with $CQ<1$. Then $PX+XQ>PX'+X'Q$.
Now this implies the inequality above in few steps: $$A_1B_1+B_1C_1+C_1D_1+D_1A_1>A_2B_1+B_1C_1+C_1D_1+D_1A_2\\ > A_2B_2+B_2C_1+C_1D_1+D_1A_2\\ > A_2B_2+B_2C_2+C_2D_1+D_1A_2\\ > A_2B_2+B_2C_2+C_2D_2+D_2A_2$$ where the first inequality holds if we set $P=D_1$, $X=A_1$, $X'=A_2$, $Q=B_1$ (and we rename the square vertices appropriately). For the second inequality set $P=A_2$, $X=B_1$, $X'=B_2$, $Q=C_1$, etc.
My picture behind the lemma is the following. If we consider the ellipse with foci $P$ and $Q$ containing $X$, then by looking at the picture it looks like that $X'$ should be inside this ellipse, which exactly means that $PX+XQ>PX'+X'Q$. I don't see a strict pure geometric agrument for this, so we can prove the lemma analytically.
Set coordinate system such that $A(-1,1)$, $B(-1,-1)$, $C(1,-1)$ and $D(1,1)$. Then $P(-1,p)$ for $0<p<1$ and $Q(1,q)$ for $-1<q<0$. It suffices to prove that $PX+XQ$ is a decreasing function when $X$ goes from $B$ to the midpoint of $[BC]$, i.e. if we set $X=(x,-1)$, it suffices to prove that $$f(x)= PX+XQ= \sqrt{(x+1)^2+(p+1)^2}+\sqrt{(x-1)^2+(q+1)^2}$$ is deacreasing on the interval $(-1,0)$. We prove that $f'(x)<0$ on this interval. $$f'(x)=\frac{x+1}{\sqrt{(x+1)^2+(p+1)^2}}+\frac{x-1}{\sqrt{(x-1)^2+(q+1)^2}}\\ =\frac{1}{\sqrt{1+(\frac{p+1}{x+1})^2}}-\frac{1}{\sqrt{1+(\frac{q+1}{x-1})^2}}.$$ The $-$ sign appeared because $-1<x<0$. So $f'(x)<0$ iff $\sqrt{1+(\frac{q+1}{x-1})^2}<\sqrt{1+(\frac{p+1}{x+1})^2}$ iff $(\frac{q+1}{x-1})^2<(\frac{p+1}{x+1})^2$ iff $(q+1)^2(x+1)^2<(p+1)^2(x-1)^2$ iff $$[(q+1)^2-(p+1)^2](x^2+1)+2[(q+1)^2+(p+1)^2]x<0.$$ The last is true because both summands are negative. The second because $x<0$ and other factors are positive, and the first because $0<p<1$ and $<-1<q<0$ imply $(q+1)^2-(p+1)^2<0$, and $x^2+1>0$.