(1) Prove that there exist five nonnegative real numbers $ a, b, c, d$ and $ e$ with their sum equal to 1 such that for any arrangement of these numbers around a circle, there are always two neighboring numbers with their product not less than $ \frac{1}{9}.$ (2) Prove that for any five nonnegative real numbers with their sum equal to 1 , it is always possible to arrange them around a circle such that there are two neighboring numbers with their product not greater than $ \frac{1}{9}.$
Problem
Source: CGMO 2003, Problem 4
Tags: inequalities unsolved, inequalities
29.12.2008 06:25
(1) Take a=b=c=1/3 and d=e=0. Clearly, two of a,b,c have to be adjacent in any arrangement, proving the claim. (2) Two of the numbers must sum to <=2/3 (or else four of the numbers sum to >4/3, contradiction). By AM-GM, these two have a product of at most 1/9, so just take an arrangement where these two #s are adjacent. (Was this really a #4 on CGMO? It seemed really easy...only took 5 min...)
24.10.2010 16:23
hmm wrote: (1) Take a=b=c=1/3 and d=e=0. Clearly, two of a,b,c have to be adjacent in any arrangement, proving the claim. (2) Two of the numbers must sum to <=2/3 (or else four of the numbers sum to >4/3, contradiction). By AM-GM, these two have a product of at most 1/9, so just take an arrangement where these two #s are adjacent. (Was this really a #4 on CGMO? It seemed really easy...only took 5 min...) I checked the original Chinese version. Part (2) should read "... such that no two neighbouring numbers have a product greater than $\frac{1}{9}$.
25.10.2010 16:22
That doesn't make it much more difficult. Assume $a\le b\le c\le d\le e$. Easy to prove that the ordering $(a,e,b,c,d)$ will do the job: As $ad\le cd,ae\le be,bc\le be$, we only have to show $be\le\frac 19$ and $cd\le\frac 19$. Suppose $cd>\frac 19$. Then by AM-GM, $c+d>\frac 23$, thus $\frac 13<d\le e$, thus $c+d+e>1$ contradiction. Suppose $be>\frac 19$, thus $3b>\frac 1{3e}$. From $3b+e\le b+c+d+e\le 1$, we get $\frac 1{3e}+e<1$ or $(e-\frac 12)^2<-\frac 1{12}$, again a contradiction.