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how can a prove that at least one of those is less than or equal to 1/4. $$\forall a,b,c\in \mathbb R^+, \ a(1-b)\leq 1/4 \lor b(1-c) \leq 1/4 \lor c(1-a) \leq 1/4.$$ help please!

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    $\begingroup$ Suppose not and multiply. $\endgroup$ – Daniel Fischer Jul 18 '14 at 22:21
  • $\begingroup$ What is the maximum value of $a(1-a)$? What about $b(1-b)$ and $c(1-c)$? Is this a contradiction? $\endgroup$ – JimmyK4542 Jul 18 '14 at 22:24
  • $\begingroup$ Then regroup - and see a contradiction. $\endgroup$ – Daniel Fischer Jul 18 '14 at 22:24
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We can assume $1-a, 1-b, 1-c \geq 0$, since otherwise we are done.

By the AM-GM inequality (see http://en.wikipedia.org/wiki/AM-GM_inequality), we have $abc(1-a)(1-b)(1-c) \leq (\frac{a+b+c+(1-a)+(1-b)+(1-c)}{6})^6= (\frac{1}{2})^6 = \frac{1}{64}$.

Then, if $a(1-b)> 1/4, b(1-c) > 1/4$ and $ c(1-a) > 1/4$, multiplying together we get $abc(1-a)(1-b)(1-c)> (\frac{1}{4})^3 = \frac{1}{64}$, which is a contradiction, and thus the result follows.

EDIT:

If you do not want to use AM-GM: Let $x \in \mathbb{R}^+ $. We have $0\leq (\sqrt x - \sqrt{(1-x)})^2 = x +(1-x) -2\sqrt{x(1-x)}$, and thus $ 2\sqrt{x(1-x)} \leq 1$, which implies $x(1-x) \leq \frac{1}{4}$. Apply this for $a,b$ and $c$, multiply together, and you get the inequality of the first paragraph.

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    $\begingroup$ What's wrong with AM-GM? It is one of the simplest inequalities out there. $\endgroup$ – JimmyK4542 Jul 18 '14 at 22:48
  • $\begingroup$ You could find the maximum of $a(1-a)$ looking at it as a quadratic polynomial in $a$. $\endgroup$ – Ragnar Jul 18 '14 at 22:50
  • $\begingroup$ @user161440 I extended my answer with an approach avoiding AM-GM, studying the function x(1-x) $\endgroup$ – Marco Flores Jul 18 '14 at 23:05
  • $\begingroup$ That's a good remark. I edited the answer: we can assume that, because otherwise, one of the three numbers we would like to be less than or equal to 1/4, would be negative, which is less than 1/4 $\endgroup$ – Marco Flores Jul 19 '14 at 17:35
  • $\begingroup$ how could you mutiply a(1-a)*b(1-b)*c(1-c) what rule is this? what if a(1-a) is negative?? $\endgroup$ – user161440 Jul 19 '14 at 17:46
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If you don't want to use AM-GM you can do it this way.

Let's asume without loss of generalization that $a\leq b\leq c$.

Trivially, If $1 \leq c$ then $b(1-c) \leq 0 \leq 1/4$.

If $a\leq 1/2 \leq b$ or $b\leq 1/2 \leq c$ then $a(1-b)\leq 1/4$ or $b(1-c)\leq 1/4$. (respectively)

If $1/2 \leq a$ then let $a' = a-1/2$ and $b' = 1/2 - (1-b) = b - 1/2$ Note that $a' \leq b'$

Then $a(1-b) = (1/2 + a')(1/2 - b') = 1/4 - (b'-a')/2 -a'b' \leq 1/4$.

Similarly, if $c \leq 1/2$. Let $a' = 1/2 - a$ and $b' = (1-b) - 1/2 = 1/2 - b$. Note that this time, $a' \geq b'$.

Then $a(1-b) = (1/2 - a')(1/2 + b') = 1/4 - (a'-b')/2 -a'b' \leq 1/4$.

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First assume $a, b, c \lt 1$.

without loss of generality, assume $a \le b$

Now consider the quadratic

$$f(x) = x^2 - x + a(1-b)$$

$f(0) = a(1-b) \gt 0$

$f(b) = b^2 - b + a(1-b) = (a-b)(1-b) \le 0$

Thus the quadratic has a real root (ok, some calculus used, but elementary proofs exist), and thus the discriminant =$ 1 -4a(1-b) \ge 0 \implies a(1-b) \le \frac{1}{4}$

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  • $\begingroup$ Am I missing a triviality that expands this for any $a,b,c\in\mathbb{R}^+$? $\endgroup$ – DanZimm Jul 19 '14 at 8:14
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    $\begingroup$ @DanZimm: If $c \ge 1$, then $b(1-c) \le 0$. $\endgroup$ – Aryabhata Jul 19 '14 at 14:53
  • $\begingroup$ Ah yes, it was a triviality, thanks! $\endgroup$ – DanZimm Jul 19 '14 at 17:31
  • $\begingroup$ For $a,b,c \le 1$, see my comment to DanZimm. There is no need of $c$ if $a \le b$. There is an implicity renaming of variables going on. For instance if you chose $a=0.3, b = 0.1, c = 0.4$, we kind of have implicitly swapped $b$ and $c$ in our proof... $\endgroup$ – Aryabhata Jul 19 '14 at 21:51

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