# Prove that $a_1+\frac{a_2^2}{a_1+a_2}+\frac{a_3^2}{a_1+a_2+a_3}>b_1+\frac{b_2^2}{b_1+b_2}+\frac{b_3^2}{b_1+b_2+b_3}.$

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?

• 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. Jun 13 '16 at 15:42
• 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. Jun 13 '16 at 15:55

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

• 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. Jun 14 '16 at 1:54