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Assuming $0 < q < 1$ and $0 < r < q$, I want to show that:

$$q^2(2-r) + q(3r^2 + 8r-4) - 8r^2 > 0 \;\text{(I)}\;\;\text{and}\;\; 3q -r-2 < 0\;\text{(II)}$$

cannot hold at the same time.

The problem seems simple enough, but so far, I can only give a very clumsy proof. Any help would be appreciated - many thanks in advance!

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Assuming $0 < q < 1$ and $0 < r < q$, show that $$q^2(2-r) + q(3r^2 + 8r-4) - 8r^2 > 0\ \ \ \tag{I}$$ $$\text{and}\;\;\;3q -r-2 < 0\tag{II}$$ cannot hold at the same time.

Strategy Hints:
Take the given constraints on $q$ and $r$,
assume BOTH inequalities hold,
derive a contradiction,
thereby proving both inequalities cannot simultaneously hold.

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The strategy is clear to me (thanks anyway). In fact, I can derive such a contradiction, but it is quite tedious (I factorize the polynomial in (I) and go from there). I'm just wondering whether a simpler argument can be given. – Martin Dec 7 '12 at 14:51
Note that $0 < r < q < 1$ so $q^2 < q, r^2 < r$. Let me know if you're still trying to tackle this. – amWhy Dec 7 '12 at 19:17
You do always a great work! – user63181 Mar 27 '14 at 16:35
up vote 2 down vote accepted

And here is an even simpler proof.

Let $f(q,r) = q^2 (2-r) + q (3r^2 + 8r-4) - 8r^2.$

It is easy to see that this function is strictly convex in $q$ over the relevant parameter range, and that $f(0,r) < 0$.

Hence there exists some unique $\tilde{q}$ such that $f(q,r) < 0$ for $q < \tilde{q}$ and $f(q,r) > 0$ for $q > \tilde{q}$ .

However, $f(q,r)$ is equal to $\frac{8}{9}(1-r)^2 (r-2) < 0$ for $q = \frac{r+2}{3}$. But due to inequality II, it cannot hold that $q \geq \frac{r+2}{3}$. It follows that both inequalities cannot hold at the same time.

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Here is an attempt for a relatively simple proof. I would be curious about your comments.

Case (a): $3r^2 + 8r- 4 > 0$ ($r > \approx 0.43$).

Then (I) is clearly increasing in $q$. But (II) implies that $q < \frac{r+2}{3}$. Hence, (I) can never attain a larger value than for $q = \frac{r+2}{3}$. Inserting this into (I), the condition

$\left(\frac{r+2}{3}\right)^2 (2-r) + \frac{r+2}{3} (3r^2 +8r-4) - 8r^2 > 0$ needs to be satisfied. But one can simplify this to

$\frac{8}{9} (1-r)^2 (r-2) > 0$, which can never hold for $0<r<1$.

Case (b): $3r^2 + 8r - 4 < 0$ ($r <\approx 0.43$).

Then it is easy to see that (I) is a strictly convex function in $q$ over the relevant range, which attains a negative value for $q=0$, first decreases, and then increases. At best, there exists some critical threshold $\tilde{q} < 1$ starting from which the function becomes positive.

But in order for (II) to be satisfied, it cannot be the case that $q \geq \frac{r+2}{3}$. However, we know from case (a) that the function in (I) is certainly negative for $q = \frac{r+2}{3}$. Thus, also case (b) both inequalities cannot hold at the same time.

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