Let $f(a,b,c)=a+\dfrac{b^2}{a+b}+\dfrac{c^2}{a+b+c}$.
Suppose $a_1 \geq a_2 \geq a_3$ and $b_1 \geq b_2 \geq b_3$ and $a_1 \geq b_1,a_2 \geq b_2,a_3 \geq b_3$.
I prove $f(a_1,a_2,a_3) \geq f(a_1,a_2,b_3) \geq f(a_1,b_2,b_3) \geq f(b_1,b_2,b_3)$.
The first inequality follows from the monotonicity of $\dfrac{x^2}{r+x}$ for any $r>0$.
The second is: if $a \geq b \geq B \geq c$, then
$\dfrac{b^2}{a^2+b}+\dfrac{c^2}{a+b+c}>\dfrac{B^2}{a^2+B}+\dfrac{c^2}{a+B+c}$,
which is equivalent to:
$\dfrac{a(b+B)+bB}{(a+b)(a+B)} \geq \dfrac{c^2}{(a+b+c)(a+B+c)}$,
which is easily seen to be true.
The third inequality is: if $a \geq A \geq b \geq c$, then $f(a,b,c) \geq f(A,b,c)$, which is equivalent to:
$1 \geq \dfrac{b^2}{(A+b)(a+b)}+\dfrac{c^2}{(a+b+c)(A+b+c)}$,
which is again seen to be true.
Each of the inequalities is tight only when the corresponding variables are equal; hence if in addition we have $a_i>b_i$ for some $i$, then we get $f(a_1,a_2,a_3)>f(b_1,b_2,b_3)$.