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Suppose that $h\in(0,1)$, $l\in(0,1)$, $p\in(0,1)$, and $l<h$. Show that:

$\frac{hl}{ph+(1-p)l}+\frac{(1-h)(1-l)}{p(1-h)+(1-p)(1-l)}<1 $

What I have done:

  1. I tried expanding the inequality. It gets very messy and I could not find any useful pattern.

  2. I tried taking derivatives with respect to $p$, $h$, and $l$ to try to spot and use any monotonicity (for example, if the LHS function strictly increased in $p$ I could prove the inequality by assuming $p=1-\epsilon$), but I could not find any pattern either.

  3. I used several numerical examples on Mathematica to confirm that the inequality holds. It seems that the assumption that $l<h$ is not even necessary, but it really does not matter whether I can relax it or not.

Any solution/idea on how this can be solved will be highly appreciated!

Thank you in advance.

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\begin{align*} &\frac{hl}{ph+(1-p)l}+\frac{(1-h)(1-l)}{\;p(1-h)+(1-p)(1-l)} < 1\\[8pt] \iff&1-\left(\frac{hl}{ph+(1-p)l}+\frac{(1-h)(1-l)}{\;p(1-h)+(1-p)(1-l)}\right) > 0\\[8pt] \iff& \frac {p(1-p)(h-l)^2}{\;\,\bigl(ph+(1-p)l\bigr)\bigl(1-(ph+(1-p)l)\bigr)} > 0\\[8pt] \end{align*} which holds since all factors of the numerator and denominator of the LHS are positive.

Explanation of positivity:

It's obvious that the factors of the numerator are positive.

For the denominator, note that $ph+(1-p)l$ is a convex combination (i.e., weighted average) of $h$ and $l$, so is at least equal to the minimum of $h$ and $l$, and is at most equal to the maximum of $h$ and $l$, hence is strictly between $0$ and $1$.

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  • $\begingroup$ Thank you so much @quasi! It took my a while to see how you managed to simplify the monster, but I see it now. Thanks :) $\endgroup$ – Lorena_dok Sep 2 at 14:12
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Idea/Hint: $ph+(1-p)l$ $(0\leq p\leq1)$is the straight line connecting h and l. Try using convexity. Edit: We differentiate twice in all variables and obtain a positive second derivative in all terms. Next, we notice the function is defined on an open cube. Due to the convexity of the functions, it takes its minimum/ maximum as each one of the variables approach the side limits, so $0$ and $1$. Just plug in all these values and compute, you will see that one can choose the variables so the function gets as close to 1 as desired, yet stays smaller.

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  • $\begingroup$ Thank you @IMOPUTFIE! I regret that my math experience is very limited. Could you help me see what you mean by using convexity? How would you advice me doing it? $\endgroup$ – Lorena_dok Sep 2 at 13:44
  • $\begingroup$ I think this might be overkill, but it definitely works. Convexity is quite a powerful mathematical tool, basically meaning the function takes its maximum/ minimum on the exterior, of a given set. (Intuitively, the connecting line between two on graph points lies higher than the graph.) $\endgroup$ – IMOPUTFIE Sep 2 at 14:05
  • $\begingroup$ I think you might be able to prove convexity more elegantly, given lines in both numerator and denominator, or by other means using this linearity(try it!), yet this (admittedly brute-force) approach definitely works. $\endgroup$ – IMOPUTFIE Sep 2 at 14:07
  • $\begingroup$ Thank you so much @IMOPUTFIE! Now I understand :) I will try to do it. $\endgroup$ – Lorena_dok Sep 2 at 14:11

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