568 Real Analysis
One possible solution to this system isa=b=c=d=^14 , in which casef(^14 ,^14 ,^14 ,^14 )=
- Otherwise, let us assume that the numbers are not all equal. If three of them are distinct,
saya, b, andc, then by subtracting the second equation from the first, we obtain
(
176
27cd−c−d)
(b−a)= 0 ,and by subtracting the third from the first, we obtain
(
176
27bd−b−d)
(c−a)= 0.Dividing by the nonzero factorsb−a, respectively,c−a, we obtain
176
27cd−c−d= 0 ,
176
27bd−b−d= 0 ;henceb=c, a contradiction. It follows that the numbersa, b, c, dfor which a minimum
is achieved have at most two distinct values. Modulo permutations, eithera=b=cor
a=bandc=d. In the first case, by subtracting the fourth equation from the third and
using the fact thata=b=c, we obtain
(
176
27a^2 − 2 a)
(d−a)= 0.Sincea =d, it follows thata=b=c=^2788 andd= 1 − 3 a= 887. One can verify that
f(
27
88
,
27
88
,
27
88
,
7
88
)
=
1
27
+
6
88
·
27
88
·
27
88
> 0.
The casea=bandc=dyields
176
27cd−c−d= 0 ,
176
27ab−a−b= 0 ,which givesa=b=c=d=^2788 , impossible. We conclude thatfis nonnegative, and
the inequality is proved.
(short list of the 34th International Mathematical Olympiad, 1993, proposed by Viet-
nam)