# Simple algebra: Show that this inequality holds

Suppose that $$h\in(0,1)$$, $$l\in(0,1)$$, $$p\in(0,1)$$, and $$l. 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 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!

\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 $$1$$.

• 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 :) – Lorena_dok Sep 2 at 14:12

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.

• 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? – Lorena_dok Sep 2 at 13:44
• 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.) – IMOPUTFIE Sep 2 at 14:05
• 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. – IMOPUTFIE Sep 2 at 14:07
• Thank you so much @IMOPUTFIE! Now I understand :) I will try to do it. – Lorena_dok Sep 2 at 14:11