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Suppose $a_1>a_2>a_3>0$ and $b_1>b_2>b_3>0$ and $a_1>b_1,a_2>b_2,a_3>b_3$.
I want to prove that
$$a_1+\frac{a_2^2}{a_1+a_2}+\frac{a_3^2}{a_1+a_2+a_3}>b_1+\frac{b_2^2}{b_1+b_2}+\frac{b_3^2}{b_1+b_2+b_3}.$$
It doesn't appear to fit nicely with any standard inequality. A brute force simplification doesn't seem to work either. Has anyone seen a clean approach to a problem like this that might work?
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 $f(a_1,a_2,a_3)>f(b_1,b_2,b_3)$.
