2
$\begingroup$

Let $a, b, c$ be three positive real numbers such that $a + b + c = 1$. Prove that among the three numbers $a − ab, b − bc, c − ca$ there is one which is at most $1/4$ and there is one which is at least $2/9$

I have proceeded by the AM-GM inequality to prove that \begin{align*}a(1 − a) \leq \frac 14, \qquad b(1 − b) \leq \frac 14, \qquad c(1 − c) \leq \frac 14.\end{align*} Multiplying these we obtain $$abc(1 − a)(1 − b)(1 − c) \leq \frac{1}{4^3}.$$ After that, I could not do much.

$\endgroup$
  • 3
    $\begingroup$ Show us your effort. $\endgroup$ – S.C.B. May 31 '16 at 16:14
  • $\begingroup$ I have proceeded by AM GM inequality to prove a(1 − a) ≤1/4 b(1 − b) ≤1/4 c(1 − c) ≤1/4 Multiplying these we obtain abc(1 − a)(1 − b)(1 − c) ≤1/4^3 after that i could not do much . $\endgroup$ – starunique2016 May 31 '16 at 16:19
  • $\begingroup$ I mean you should add that information to the question, not in a comment. $\endgroup$ – S.C.B. May 31 '16 at 16:20
0
$\begingroup$

Hint 1: For the first inequality, without loss of generality let $a$ be the smallest of the three numbers. Then by the AM-GM, we have $$a(1-b) \le \left(\frac{a+(1-b)}2\right)^2.$$

Hint 2: For the second inequality, let $a$ be the largest number and $b$ the smallest. Then $a \ge \frac13$ and $1-b \ge \frac23$.

$\endgroup$

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.