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 $b$.

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

$\endgroup$
  • $\begingroup$ How did you come up with the $\frac{95}{3}-40a$ factor? It doesn't look very intuitive :) $\endgroup$ – pi66 Aug 19 '16 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 '16 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 '16 at 5:33
  • $\begingroup$ @mathlove Good catch! I'll fix it in a moment. $\endgroup$ – timon92 Aug 20 '16 at 7:21
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Not sure how this helps.. $\frac{1-a^2}{3-2a}$ is neither an increasing nor a decreasing function. $\endgroup$ – pi66 Aug 15 '16 at 12:23
  • $\begingroup$ you must do case work $\endgroup$ – Dr. Sonnhard Graubner Aug 15 '16 at 12:23
  • $\begingroup$ That looks like being really messy. I hope there's a better way $\endgroup$ – pi66 Aug 15 '16 at 12:37
  • $\begingroup$ you must consider the cases $$a\geq b$$ or $$a<b$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 15 '16 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.