Re: x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 <= C(x1^2 + x2^2 + x3^3 + x4^2)^{3/2}

On Wed, 17 Aug 2005 22:38:01 GMT, Kira Yamato <kirakun@xxxxxxxxxxxxx>
wrote:

>I've having trouble showing for any four real numbers x1, x2, x3, x4,
>there exists a real constant C such that
>
>x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 <= C(x1^2 + x2^2 + x3^3 + x4^2)^{3/2}.
>
>This is what I got so far: We can reduce it to the case of
>considering the unit-sphere only. This is because the equation is
>homogeneous. So, next I try to use lagrange multiplier to find the
>maximum value of the left-hand side on the unit-sphere. However, the
>algebra becomes unwielding and I cannot proceed.
>
>Now this was an in-class exam question, and it should not have required
>more than 15 minutes to solve. Am I missing a trick that I don't see?
>
>I also tried using symmetry argument. The left-side side is symmetry
>for every pair of coordinates. So, it's intuitive to guess that the
>maximum occurs at x1=x2=x3=x4. However, I don't know how to justify
>this rigorously.
>
>Please help. Thanks.
>
>-kira

Firstly, I suspect you are not stating the problem correctly. Look at
the following 5 problems (the first is the one you gave):

(1) Show that for any four real numbers x1, x2, x3, x4, there exists
a real constant C such that the following inequality holds:x1x2x3 +
x1x2x4 + x1x3x4 + x2x3x4 <= C(x1^2 + x2^2 + x3^3 + x4^2)^{3/2}.

(2) Show that there exists a real constant C such that for any four
real numbers x1, x2, x3, x4, the following inequality holds:
x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 <= C(x1^2 + x2^2 + x3^3 +
x4^2)^{3/2}.

(3) Find a real constant C such that for any four real numbers x1,
x2, x3, x4, the following inequality holds:
x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 <= C(x1^2 + x2^2 + x3^3 +
x4^2)^{3/2}.

(4) Show that there exists a least real constant C, such that for any
four real numbers x1, x2, x3, x4, the following inequality holds:
x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 <= C(x1^2 + x2^2 + x3^3 +
x4^2)^{3/2}.

(5) Show that there exists a least real constant C, and then find the
value of C, such that for any four real numbers x1, x2, x3, x4, the
following inequality holds:
x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 <= C(x1^2 + x2^2 + x3^3 +
x4^2)^{3/2}.

For problem (1), look at the order of the quantifiers. You have the
phrase "for any four real numbers" before the phrase "there is a real
constant C". Taken literally, problem (1) is actually too easy. The
point is that you don't need a universal C, all you need is a C for
each choice of x1,x2,x3,x4. So suppose x1,x2,x3,x4 are given.
Consider 2 cases ...
case 1: x1^2+x2^2+x^3^2+x4^2=0
In this case, x1=x2=x3=x4=0 so you can choose any C and you have
equality.
case 2: x1^2+x2^2+x3^2+x4^2>0,
In this case, simply choose C=(x1x2x3 + x1x2x4 + x1x3x4 +
x2x3x4)/(x1^2 + x2^2 + x3^3 + x4^2)^{3/2}, and again you have
equality.
Thus, in both cases you can achieve equality (hence also <=), so you
have solved the problem. Of course, the problem was made trivial since
the wording made it allowable to choose a value of C based on the
values of x1,x2,x3,x4, as opposed to a single value of C that would
for all x1,x2,x3,x4.

More than likely, the problem you were given was not problem (1), but
one of the other problems -- either (2),(3),(4), or (5+.

Notice the subtle difference in problems (2), (3), (4), (5).

Problem (2) asks you to show there exists a C. You don't have to
actually find one, and you don't need to find a least such C.

Problem (3) asks you to actually find a C, but it's not required to be
the least.

Problem (4) only asks you to show that a least C exists, but still
doesn't ask you to find it.

Problem (5) asks you to show a least C exists and then also asks you
to find it.

So look at problem (2).

You still have to worry about the special case x1=x2=x3=x4=0, but as
before, any value of C will work in that case, so deal with that
later. If the x1,x2,x3,x4 are not all zero, then the homogeneity
property of the equation, which you recognized, does allow you to
assume x1^2+x2^2+x3^2+x4^2=1, and that was a good observation since it
dramatically simplifies the problem. Let S denote the unit sphere in
R^4 and Let c(x)=x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4. So, assuming
x=(x1,x2,x3,x4) is on S, the inequality being analyzed reduces to
c(x)<=C. Note that any upper bound for c(x) on S gives a value of C
and conversely, any value of C which works for all point x on S is an
upper bound for c(x). So for problem (2), we only need to show that
such an upper bound exists. Now c(x) is clearly continuous on all R^4,
so it's certainly continuous on S. Also S is compact (closed and
bounded), hence, since any continuous function on a compact set is
bounded, it follows that c(x) has an upper bound on S, therefore a
value of C exists. As noted before, such a C also works for
x1=x2=x3=x4=0, so problem (2) is resolved.

Next look at problem (3).

Here we need to actually produce a C, but it doesn't have to be best
possible. Ok, then just look at each term of c(x). Can you see an easy
upper bound for each term? Well, since x is on S, each coordinate is
in the interval [-1,1]. Then each term of c(x) is bounded above by 1,
so c(x) is bounded above by 4. Thus, for problem (3), we can take C=4,
and of course, C=4 works for the case x=0 as well. Note that we are
just answering problem (3) by finding some C for which the inequality
is always satisfied. Is C best possible? Almost certainly not, but for
problem (3), especially under test conditions, who cares?

Next, problem (4)..

To resolve problem (4), use the same argument as for problem (2), but
note that since c(x) is continuous on the compact set S, c(x) is not
only bounded but actually attains both a minimum value and a maximum
value on S. The maximum value of c(x) can be used for C, which shows
that a least value of C exists. Again, the case x1=x2=x3=x4 is not
picky -- the same C works. This resolves problem (4).

Finally look at problem (5).

To show that a least value of C actually exists, argue as in problem
(4). So now we have to actually find this least value, or
equivalently, we need to find the maximum value of c(x) on S.

In some classes, they don't bother requiring you to prove that a least
C exists -- they simply let you assume there is a least such C and
then ask you to find it. But be careful -- I've seen problems where a
function f(x) has no maximum, but the student, unaware of the issue,
applies the method of Lagrange multipliers, which finds local extrema.
The student, oblivious to the trap, proceeds to evaluate f at all the
local extrema, takes the maximum of those values, and then asserts
that it's the global maximum value of f.

Ok, but we've been careful -- we've shown that c(x) has a maximum on
S, so now we only need to find it.

I have to admit, I didn't really try the Lagrange multiplier method on
this problem until now, since I was concentrating on explaining the
easier problems (1), (2), (3), (4). But now that I've applied the
method, it doesn't look so promising. I get 5 equations, 5 unknowns
(x1,x2,x3,x4,lambda), but the system is nonlinear in all variables,
and I don't immediately see a trick that can be used to solve.

But there are alternatives. Let's recall your claim that by symmetry,
the maximum will occur for x1=x2=x3=x4. That may be good intuition
based on experience, but it's not a proof that the maximum value
occurs there, it's just a likely place. Still, it does suggest a
method of attack.

First, we can argue that the maximum value can be achieved with all
nonnegative coordinates, since by replacing any x=(x1,x2,x3,x4) on S
with the point y=(|x1|,|x2|,|x3|,|x4|), y stays on S and c(x)<=c(y)
(c(x)<=c(y) term by term). So we will restrict our attention to
nonnegative points of S.

Suppose we can show that given any nonnegative point x=(x1,x2,x3,x4)
on S such that two coordinates are unequal, x1,x2, suppose, then c(x)
is increased by replacing (x1,x2) by (u,u) where u is chosen so that
x1^2+x2^2=2u^2 and u>=0. The choice of u insures that the new point
x'=(u,u,x3,x4) is still on S and is still nonnegative. So now if we
let x=(x1,x2,x3,x4) be a nonnegative point on S at which the c(x) is
maximized, then since c(x)<c(x') is impossible, we must have x1=x2,
and by analogous arguments, every pair of coordinates must be equal.
Hence x1=x2=x3=x4, which, since we have nonnegativity, gives a unique
point on x on S and we can then simply evaluate c(x) at that point to
get the maximum value of c(x) on S. The maximum thus obtained is then
the least value of C for which the original inequality is always
satisfied.

Note that this argument depends on knowing that c(x) actually has a
maximum, but we resolved that in problem (4).

So it remains to show that if x1 and x2 are unequal, then
c(x1,x2,x3,x4) < c(u,u,x3,x4) where u is the nonnegative constant so
that x1^2+x2^2=2*u^2. Since x1,x2 are unequal, they are not both zero,
so u>0. Now keep x3,x4 fixed (regard them as constant) and let x1,x2
vary subject to x1^2+x2^2=2*u^2 (keeping u constant). Now you can use
the method of Lagrange multipliers on c(x) regarded now as a function
of just 2 variables (x1,x2) subject to the constraint x1^2+x2^2=2*u^2
(which is the locus of a circle in R^2). Without working out the
details, it's easy to see that the new equations produced by the
Lagrange multiplier method in this restricted setting are linear and
so are easy to solve. I'm confident that when you work it out, it will
show that the maximum occurs when x1=x2 and then, since x1,x2,x3,x4
are all nonnegative, the unique maximum occurs at (u,u), showing that
c(x1,x2,x3,x4) < c(u,u,x3,x4) as was claimed.

This took longer than I expected -- sorry, but that's because I chose
to analyze various alternative statements of the problem and explore
some of the issues, as if from the student's perspective.

quasi
.

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