2
$\begingroup$

Let $0\leq a,b\leq 1$. Prove that $$\max\left(\frac{1-a^2}{3-2a},\frac{1-b^2}{3-2b}\right)\geq\frac{12}{19}\cdot\frac{3-a^2-2b^2}{6-a-3b}.$$

Equality holds when $a=0$ and $b=2/3$. This seems to be the only case of equality.

The left-hand side is symmetric in $a$ and $b$ but the right-hand side isn't, so we can't simply assume $a\geq b$ or vice-versa.

A natural thing to try would be to divide into two cases according to which term in the max is greater, but the inequality $$\frac{1-a^2}{3-2a}\geq \frac{1-b^2}{3-2b}$$ doesn't translate easily to nice form of $a$ in terms of $b$.

$\endgroup$
1
  • $\begingroup$ since the left side is symmetric and max commutative, the inequality must hold with a swap on the right $\endgroup$
    – user354674
    Aug 18, 2016 at 16:19

2 Answers 2

1
+50
$\begingroup$

Lemma. If $x,z\ge 0$, $y,t,\lambda>0$, and $\displaystyle \frac xy , \frac zt < \lambda$ then $\displaystyle \frac{x+z}{y+t}<\lambda$.

Proof of lemma: $$\frac{x+z}{y+t} < \frac{\lambda y + \lambda t}{y+t} = \lambda.$$

For the sake of contradiction, assume that both $\displaystyle \frac{1-a^2}{3-2a}$ and $\displaystyle \frac{1-b^2}{3-2b}$ are less than $\displaystyle \frac{12}{19} \cdot \frac{3-a^2-2b^2}{6-a-3b}$.

Putting $x=1-a^2$, $y=3-2a$, $z=2-2b^2$, $t=6-4b$, $\lambda=\displaystyle \frac{12}{19} \cdot \frac{3-a^2-2b^2}{6-a-3b}$ in the lemma we get $$\frac{3-a^2-2b^2}{9-2a-4b} < \frac{12}{19} \cdot \frac{3-a^2-2b^2}{6-a-3b}.$$ The nominators cancel out. Simplifying yields $6+5a<9b$.

Therefore, if we additionally assume at the beginning that $6+5a\ge 9b$ then we reach a contradiction and therefore the inequality is proved in this case.

In the case $6+5a<9b$ we have $\displaystyle a<\frac 35$ and $\displaystyle b>\frac 23$. From here we get $$\frac{1-a^2}{3-2a} \ge \frac 13 \ge \frac{1-b^2}{3-2b}.$$ Therefore we need to prove that $$\frac{1-a^2}{3-2a} \ge \frac{12}{19} \cdot \frac{3-a^2-2b^2}{6-a-3b}.$$ Clearing denominators and simplifying we get an equivalent form $$72b^2-48ab^2+57a^2b-57b+6+53a-78a^2-5a^3 \ge 0$$

which is true because \begin{align} &72b^2-48ab^2+57a^2b-57b+6+53a-78a^2-5a^3 \ge \\ \ge \ &72b^2-48ab^2+57a^2b-57b+6+53a-78a^2+a^2(6-9b) = \\ =\ &(3b-2)\left(-3-\frac{32}{3}a+16a^2+8(3-2a)b\right) + a\left(\frac{95}{3}-40a\right) \ge \\ \ge \ &(3b-2)\left(-3-\frac{32}{3}a+16a^2+8(3-2a)\frac{6+5a}{9}\right) + a\left(\frac{95}{3} - 40\cdot \frac 35 \right) = \\ =\ &(3b-2)\left(13-8a+\frac{64}{9}a^2 \right) + \frac{23}{3}a \ge \\ \ge \ &0. \end{align}

$\endgroup$
4
  • $\begingroup$ How did you come up with the $\frac{95}{3}-40a$ factor? It doesn't look very intuitive :) $\endgroup$
    – pi66
    Aug 19, 2016 at 12:23
  • $\begingroup$ I just divided the polynomial by $(3-2b)$ and the remainder was $a(\frac{95}3 - 40a)$ $\endgroup$
    – timon92
    Aug 19, 2016 at 12:42
  • 1
    $\begingroup$ Seems wrong because $72b^2-48ab^2+57a^2b-57b+6+53a-78a^2+a^2(6-9b) =$$ (3b-2)\left(-3-\frac{32}{3}a+4a^2+8(3-2a)b\right) + a\left(\frac{95}{3}-40a\right)$ does not hold in general. $\endgroup$
    – mathlove
    Aug 20, 2016 at 5:33
  • $\begingroup$ @mathlove Good catch! I'll fix it in a moment. $\endgroup$
    – timon92
    Aug 20, 2016 at 7:21
0
$\begingroup$

hint: $$\frac{1-a^2}{3-2a}-\frac{1-b^2}{3-2b}=\frac{(a-b)(2-3a-3b+2ab)}{(2a-3)(2b-3)}$$

$\endgroup$
4
  • $\begingroup$ Not sure how this helps.. $\frac{1-a^2}{3-2a}$ is neither an increasing nor a decreasing function. $\endgroup$
    – pi66
    Aug 15, 2016 at 12:23
  • $\begingroup$ you must do case work $\endgroup$ Aug 15, 2016 at 12:23
  • $\begingroup$ That looks like being really messy. I hope there's a better way $\endgroup$
    – pi66
    Aug 15, 2016 at 12:37
  • $\begingroup$ you must consider the cases $$a\geq b$$ or $$a<b$$ $\endgroup$ Aug 15, 2016 at 12:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .