ELMO 2014 SL A1
by Wolstenholme, Aug 1, 2014, 9:36 PM
In a non-obtuse triangle 
, prove that
.
Proof:
Using the substitution
it suffies to show that if 
then 
.
I shall use the method of Lagrange Multipliers to show this.
Define
and 
. Note that 
and 
and all of their partial derivatives are continuous. Moreover, note that 
at all points since at most one of 
can equal 
.
Now, define the open set:
![\[ U =\left\{ (a,b,c)\mid 0 < a,b,c < 1000 \right\}. \]](//latex.artofproblemsolving.com/3/4/c/34c1b17d509b5ae9db21b3c453a542736da808bd.png)
Its closure is:
![\[ \overline U =\left\{ (a,b,c)\mid 0 \leq a,b,c \leq 1000 \right\}. \]](//latex.artofproblemsolving.com/3/1/c/31ce4d702229bf4f29ef4452fae653fff87872c4.png)
Therefore the constraint set:
Is compact, and so obtains a global maximum over this set.
Now if
lies on the boundary, one of must clearly be equal to (if one of is equal to 
the claim is trivial). In this case, it suffices to show that if 
then 
but this is trivial. Now, we assume 
and consider a local maximum.
Then we must have
equal to 
for some real 
.
Now let
. We have 
equal to 
. Since 
and 
, subtracting these two equations yields that either 
or that 
. For now, assume that 
. Then 
which, upon expanding and dividing both sides by 
, implies that 
. But because we know that 
. Therefore we must have 
, absurd. Therefore at least two of 
must be equal, which implies that at least two of are equal.
Assume WLOG that
. It suffices to show that if 
then 
. Solving for 
in terms of 
, substituting and clearing denominators it suffices to show that if 
then 
. So it suffices to show that 
. But since 
, we are done.
Phew!
, prove that
.Proof:
Using the substitution
it suffies to show that if 
then 
.I shall use the method of Lagrange Multipliers to show this.
Define
and 
. Note that 
and 
and all of their partial derivatives are continuous. Moreover, note that 
at all points since at most one of 
can equal 
.Now, define the open set:
![\[ U =\left\{ (a,b,c)\mid 0 < a,b,c < 1000 \right\}. \]](http://latex.artofproblemsolving.com/3/4/c/34c1b17d509b5ae9db21b3c453a542736da808bd.png)
Its closure is:
![\[ \overline U =\left\{ (a,b,c)\mid 0 \leq a,b,c \leq 1000 \right\}. \]](http://latex.artofproblemsolving.com/3/1/c/31ce4d702229bf4f29ef4452fae653fff87872c4.png)
Therefore the constraint set:
\[ \overline S =\left\{\mathbf{x}\in\ol \overline U : g(\ol \mathbf{x}) = 1\right\} \]Is compact, and so
obtains a global maximum over this set.Now if
lies on the boundary, one of must clearly be equal to (if one of is equal to 
the claim is trivial). In this case, it suffices to show that if 
then 
but this is trivial. Now, we assume 
and consider a local maximum.Then we must have
equal to 
for some real 
.Now let
. We have 
equal to 
. Since 
and 
, subtracting these two equations yields that either 
or that 
. For now, assume that 
. Then 
which, upon expanding and dividing both sides by 
, implies that 
. But because we know that 
. Therefore we must have 
, absurd. Therefore at least two of 
must be equal, which implies that at least two of are equal.Assume WLOG that
. It suffices to show that if 
then 
. Solving for 
in terms of 
, substituting and clearing denominators it suffices to show that if 
then 
. So it suffices to show that 
. But since 
, we are done.Phew!
This post has been edited 1 time. Last edited by Wolstenholme, Aug 1, 2014, 11:10 PM
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