Monotone Increasing Concave Function

Question

Given monotone increasing concave function $f(x):\mathcal{R}_{\geq 0} \to \mathcal{R}_{\geq 0}$, Can we say that $$ f(d_1)+f(d_2)-f(d_1+d_2) \leq f(d_3)+f(d_4)-f(d_3+d_4) $$

if $d_1<d_3$ and $d_2<d_4$?

Answer 1

I think it's true for all concave function $f:\mathbb R^+\rightarrow\mathbb R^+$.

Let $d_1=a$, $d_2=b$, $d_3=c$ and $d_4=d$.

Thus, $a<c$, $b<d$ and we need to prove that $$f(d)+f(c)+f(a+b)\geq f(d+c)+f(a)+f(b).$$ We can assume that $d=\max\{a,b,c,d\}$.

Consider three cases.

1. $a<c\leq b<d.$

Since $c+d>b>a$ and for $d>c\geq a+b$ we have $$(c+d,b,a)\succ(d,c,a+b),$$ by Karamata we obtain: $$f(d)+f(c)+f(a+b)\geq f(c+d)+f(b)+f(a).$$ Also, for $d\geq a+b\geq c$ we have $$(c+d,b,a)\succ(d,a+b,c),$$ which by Karamata again gives $$f(d)+f(a+b)+f(c)\geq f(c+d)+f(b)+f(a).$$ If $a+b\geq d>c$ then $$(c+d,b,a)\succ(a+b,d,c)$$ and by Karamata we obtain: $$f(a+b)+f(d)+f(c)\geq f(c+d)+f(b)+f(a).$$ 2. $a\leq b\leq c<d.$

In this case the proof is the same.

3. $b\leq a<c<d.$

In this case $$(c+d,a,b)\succ(d,c,a+b)$$ if $d>c\geq a+b,$ $$(c+d,a,b)\succ(d,a+b,c)$$ if $d\geq a+b\geq c$, $$(c+d,a,b)\succ(a+b,d,c)$$ if $a+b\geq d>c$ and we get our inequality by Karamata.