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Suppose that, we have some real numbers $r_1,r_2,\dots,r_m \in [0,1]$; and we study a sum, $$ S_1= \sum_{i=1}^m \binom{m}{i}(-1)^{i-1}f(r_i), $$ for $f:[0,1]\to[0,1]$ a concave bijection. Now, take another sequence $z_1,\dots,z_n\in [0,1]$ with $r_i\geq z_i$ for $i=1,2,\dots,m$.

Can we prove or disprove, $$ S_1 = \sum_{i=1}^m \binom{m}{i}(-1)^{i-1}f(r_i) \geq S_2 = \sum_{i=1}^m \binom{m}{i}(-1)^{i-1}f(z_i)? $$ In case, does it depend on the very function $f(\cdot)$ (for instance, can we say something, for $f(x)=x(2-x)$). Also, for $f(x)=x$.

Apparently I've overly-generalized this. I simply need; $f(x)=x$ , $z_i=r_i^2,\forall i$. In fact I have some monotonicity between $r_i$'s; but I guess that is a bit irrelevant ($1\geq r_1 \geq r_2 \geq \cdots \geq r_m\geq 0$).

I tried to pull some results from majorization; but the presence of annoying signs makes it somehow untouchable. Any help is appreciated!

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  • $\begingroup$ A simple numerical experiment throws up a lot of false instances of your inequality for $f(x)=x(2-x)$ for $m=2,3,\ldots,10.$ $\endgroup$
    – kodlu
    Commented May 13, 2018 at 23:36
  • $\begingroup$ (removed old comment). kodlu; thanks for the answer. I've somehow ill-phrased the problem; and overly-generalized it. I need $f(x)=x$; $z_i=r_i^2$ (simply, the first sum with $r_i$'s and the second with $r_i^2$'s). $\endgroup$
    – anonymous_man
    Commented May 13, 2018 at 23:58
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    $\begingroup$ Certainly not: take $r_i=z_i$ for every $i\ne2$ and take $r_2 > z_2$; then $S_1-S_2 = -\binom m2 (r_2-z_2) < 0$. $\endgroup$
    – Greg Martin
    Commented May 14, 2018 at 0:11
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    $\begingroup$ Take $m=2$, $r_1=1$, $r_2=0.5$. $\endgroup$
    – fedja
    Commented May 14, 2018 at 1:58

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