1984 IMO Problems/Problem 1
Problem
Let ,
,
be nonnegative real numbers with
. Show that
Solution 1
Note that this inequality is symmetric with x,y and z.
To prove note that
implies that at most one of
,
, or
is greater than
. Suppose
, WLOG. Then,
since
, implying all terms are positive.
To prove , suppose
. Note that
since at most one of x,y,z is
. Suppose not all of them equals
-otherwise, we would be done. This implies
and
. Thus, define
,
Then,
,
, and
. After some simplification,
since
and
. If we repeat the process, defining
after similar reasoning, we see that
.
Solution 2
By the method of Lagrangian multipliers. Let and
. We will find the local maxima/minima of
over
subject to
.
Let .
is bounded and the intersection of two closed sets, hence compact. So every sequence in
has a convergent subsequence. Hence if
is such that
converges to the infimum or supremum of
in
, there will be a convergent subsequence
and
by the continuity of
. Hence the infimum/supremum of
over
will also be a local minima/maxima of
over
.
We must solve . This is equivalent to
. If
then
. WLOG
and we have
.
Then WLOG
. These imply
. Then
since
. We get
.
If then letting
one gets
which imply
since
. This implies
since
. And
.
Now to check the boundary of . WLOG we must consider the cases
and
. If
then
by
so
. If
substituting
in
yields
which is between
and
since
. Considering all the values found we find
.
~not_detriti
Video Solution
Video Solution
See Also
1984 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |