$a_1 , a_2, a_3 , a_4 , b_1 , b_2 , b_3 , b_4\in\Bbb R , p\in(0, 1)$. $a_1\leq a_2\leq a_4 , a_1\leq a_3\leq a_4 , b_1\leq b_2\leq b_4 , b_1\leq b_3\leq b_4 $.

Show that $$ a_1b_1(1-p)^2+(a_2b_2+a_3b_3)p(1-p)+a_4b_4p^2 \geq\Big[a_1(1-p)^2+(a_2+a_3)p(1-p)+a_4p^2\Big]\Big[b_1(1-p)^2+(b_2+b_3)p(1-p)+b_4p^2\Big]$$

I want to show that LHS- RHS $\geq0$. But

LHS- RHS = $ p(1-p)[(-a_1+a_2+a_3-a_4)(-b_1+b_2+b_3-b_4)p^2+...]$

It's difficult to move forward.

Thank you very much

  • $\begingroup$ Are you sure this is true? Where did it come from? If the a's are 1,2,3,5 and the b's are 1,2,3,3.1 then your product for LHS-RHS is negative. $\endgroup$ Aug 28 '15 at 6:24
  • $\begingroup$ @martycohen $p(1-p)(-0.9p^2-0.6p+5)\geq0$ $\endgroup$
    – Clin
    Aug 28 '15 at 9:09

There is a mechanical process to solve the question, however it's so tedious that I never finish the whole process. (tscjlt99 solved it by Mathematica 8.0) We do some substitutions as follows:


$b_2=b_1+qy$,$b_3=b_1+ry$,$b_4=a_1+y$, where $0\leq p,q,r,s,t\leq 1$ and $x,y\geq 0$

Then $LHS-RHS=p(1-p)G(p,q,r,s,t)xy\geq0\Leftrightarrow G(p,q,r,s,t)\geq 0$, where


$$\frac{\partial G}{\partial i}=0$$, where $i=p,q,r,s,t$

The solution is $(p,q,r,s,t)=(1/2,1/2,1/2,1/2,1/2)$

From $G(1/2,1/2,1/2,1/2,1/2)=1/2>0$, we only need to prove the following ten inequalties to finish the proof. $G(1,q,r,s,t)\geq 0$, $G(p,1,r,s,t)\geq 0$, $G(p,q,1,s,t)\geq 0$, $G(p,q,r,1,t)\geq 0$, $G(p,q,r,s,1)\geq 0$, $G(0,q,r,s,t)\geq 0$, $G(p,0,r,s,t)\geq 0$, $G(p,q,0,s,t)\geq 0$, $G(p,q,r,0,t)\geq 0$, $G(p,q,r,s,0)\geq 0$.

The question is degraded from 1 inequality with five variables to 10 inequalities with four variable. Probably, these ten inequalities are very simple to prove, if not, We can proceed this process until the question is degraded to inequalities with only 1 variable.

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