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


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?

  • $\begingroup$ The function $\dfrac{x^2}{c+x}$ is increasing for $c>0$, so you can assume that $b_3=a_3$ - I know that $b_3<a_3$, but you can prove things with $\geq$ and later look at when equality happens. $\endgroup$
    – Aravind
    Jun 13, 2016 at 15:42
  • $\begingroup$ That is, I am suggesting the path: $f(a_1,a_2,a_3) \geq f(a_1,a_2,b_3) \geq f(a_1,b_2,b_3) \geq f(b_1,b_2,b_3)$ since this is both necessary and sufficient (assuming $a_1 \geq b_1$ etc and later make the inequalities strict.) In the previous comment, I point out that the first inequality is true. $\endgroup$
    – Aravind
    Jun 13, 2016 at 15:55

1 Answer 1


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 $f(a_1,a_2,a_3)>f(b_1,b_2,b_3)$.

  • $\begingroup$ Thanks for the input! I was worried that an inductive type answer might be the only answer. Do you suspect that this is the case? I'll accept the answer if nothing more elegant is suggested. $\endgroup$
    – Auslander
    Jun 14, 2016 at 1:54

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