Symmetric Polynomial Inequalities in Three Variables
by MathPassionForever, Sep 3, 2019, 1:30 PM
Hey guys! I shall present here, contrary to what you would normally expect me to, some strong INEQUALITY STUFF! Here we go:
We'll start off with with something which has REALLY ugly looks, but get along with it, it gets MUCH nicer in the end.
Define
Theorem 1: Given
, there exist corresponding 
iff:
![\begin{align*}
&(i)u^2 \geq v^2\\
&(ii)w \in\left[3uv^2 - 2u^3 - 2\sqrt{(u^2 - v^2)^3}, 3uv^2 - 2u^3 + 2\sqrt{(u^2-v^2)^3}\right]
\end{align*}](//latex.artofproblemsolving.com/a/e/8/ae839bfe7a2c3092664568d61663cd641da814e3.png)
Proof:
Define
. Clearly the roots of this equation are 
.
Lemma 1:
One way is clear. The other way is the one we need to focus on.
Let
and let it denote the complex number 
. Since the coefficients of 
are real, there must be another nonreal root, WLOG 
, which is 
. Plug into our expression to get 
.
After our lemma, the problem looks like this:
Given
, there exist corresponding iff 
I'll skip some computational pain for you and present right away that thing thing becomes
Condition (i) clearly is necessary here (It is of course necessary, but here we learn why it is part of criteria that make it sufficient). And Condition (ii) is mere rewriting of what I just presented!
Theorem 2:
Proof: One direction is obvious, the other direction is what I shall prove i.e. if at least one of is negative, then so is the case for 
.
The case when all three are negative gives us negative
.
The case where one is negative gives negative
.
Now we assume that exactly two are negative, say
. Also assume that 
is nonnegative i.e. 
.Now, 
, and we're done.
Okay, this was quite boring so far. But, just as I said, stay with me, please.
Hey, have you noticed that in most positive real inequalities, equality cases are where three, or more generally speaking, two variables are equal, or one of them is zero? Well, the following theorem explains why!
Tej's theroem: There are three parts of this theorem. I'll write all the three statements, but present the proof to just one, for each of them has nearly the same proof. Well, let me get straight to the point. That's an exercise for the readers.
(i) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former is attained only when two of are equal to each other and the latter is attained only when two of are equal or one of is zero.
(ii) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.
(iii) Fix
and let there exist a value of such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.
I shall prove only the former.
We know that. Now substitute 
and write it in the form of a polynomial of 
, i.e. 
. While 
and 
can be hideous, what is important is that 
.So as tends to infinity, 
is definitely negative. And we assumed that there exists a value of for which is positive. This means that has a positive root. Let 
be the largest root and 
be the smallest positive root. Equality is clearly achieved iff the polynomial is zero i.e. 
or 
, i.e.
. Take your time to realise this.
Aha! this powerful theorem gives us the nuke we were waiting for! * Insert devious laughter *
SD-3,4,5 theorem(For all nonnegative
tho).
To verify if a symmetric homogeneous 3 variable polynomial
of degree is nonnegative for nonnegative , we just need to verify the cases:
(i)
(ii)
Told you we'd get to nice stuff
Proof:
Every polynomial function in 3 variables of degree
can be written as 
. Fixing we maximise or minimise to find extrema of function. And extrema are achieved when two variables are equal or one is zero, and we're done!
Based on similar logic, just that you need to focus on the Tej's theorem reasoning,to find this:
Let be sides of a possibly degenerate triangle instead. Then we need to verify this just for:
(i)![$f(x,1,1) \geq 0 \forall x \in [0,2]$](//latex.artofproblemsolving.com/3/9/e/39eaeeae6b9500155245b4c1cf1ba844777a2ea4.png)
(ii)
We also have a non-homogenous variant!
Non-homogenous variant: Let satisfy all criteria that it did earlier, except homogeneity. We can still check for just the cases:
(i)
(ii)
Some crucial facts that help you do manipulations as per your need:
(i)If the degree is
, you can make it a polynomial in instead and check that 
only from the second part of Tej's theorem!
(ii) We used the fact that has 
as functions. But they can be ANY functions, not just polynomials!
Edit: Hey guys, back with part of what I promised!
SD-6 Theorem: Let be degree 6 instead (and follows all other criteria as mentioned earlier). Then, it can be written as
![\[ f(a,b,c) = Aw^6 + B(u,v^2)w^3 + C(u,v^2) \]](//latex.artofproblemsolving.com/c/9/9/c992e31cbd796e035d9da84c0f2f439bc9688876.png)
After noting that 
is a constant, see that there are two cases:
(i)
: Just check if and . The same logic as used in Tej's theorem.
(ii) Um..... this is the bad part. For
, we need to prove
![\[ f(a,b,c) \geq A\left(w^3 + Ku^3 + Luv^2 + M\dfrac{v^4}{u}\right)^2 \]](//latex.artofproblemsolving.com/8/c/d/8cd30bcc953376100064a7fa43022afad049259b.png)
So yeah, we need to find 
ourselves. The point of doing this is to make 
to apply (i) now.
Adding problems soon!
Dec 27:
Edit: Ok maybe not so soon
April 1:
Edit: Can we act like I never said this?
We'll start off with with something which has REALLY ugly looks, but get along with it, it gets MUCH nicer in the end.
Define
Theorem 1: Given
, there exist corresponding 
iff:![\begin{align*}
&(i)u^2 \geq v^2\\
&(ii)w \in\left[3uv^2 - 2u^3 - 2\sqrt{(u^2 - v^2)^3}, 3uv^2 - 2u^3 + 2\sqrt{(u^2-v^2)^3}\right]
\end{align*}](http://latex.artofproblemsolving.com/a/e/8/ae839bfe7a2c3092664568d61663cd641da814e3.png)
Proof:Define
. Clearly the roots of this equation are 
.Lemma 1:
One way is clear. The other way is the one we need to focus on.
Let
and let it denote the complex number 
. Since the coefficients of 
are real, there must be another nonreal root, WLOG 
, which is 
. Plug into our expression to get 
.After our lemma, the problem looks like this:
Given
, there exist corresponding iff 
I'll skip some computational pain for you and present right away that thing thing becomes
Condition (i) clearly is necessary here (It is of course necessary, but here we learn why it is part of criteria that make it sufficient). And Condition (ii) is mere rewriting of what I just presented!
Theorem 2:
Proof: One direction is obvious, the other direction is what I shall prove i.e. if at least one of
is negative, then so is the case for 
.The case when all three are negative gives us negative
.The case where one is negative gives negative
.Now we assume that exactly two are negative, say
. Also assume that 
is nonnegative i.e. 
.Now, 
, and we're done.Okay, this was quite boring so far. But, just as I said, stay with me, please.
Hey, have you noticed that in most positive real inequalities, equality cases are where three, or more generally speaking, two variables are equal, or one of them is zero? Well, the following theorem explains why!
Tej's theroem: There are three parts of this theorem. I'll write all the three statements, but present the proof to just one, for each of them has nearly the same proof. Well, let me get straight to the point. That's an exercise for the readers.
(i) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former is attained only when two of are equal to each other and the latter is attained only when two of are equal or one of is zero.(ii) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.(iii) Fix
and let there exist a value of such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.I shall prove only the former.
We know that
. Now substitute 
and write it in the form of a polynomial of 
, i.e. 
. While 
and 
can be hideous, what is important is that 
.So as tends to infinity, 
is definitely negative. And we assumed that there exists a value of for which is positive. This means that has a positive root. Let 
be the largest root and 
be the smallest positive root. Equality is clearly achieved iff the polynomial is zero i.e. 
or 
, i.e.
. Take your time to realise this.Aha! this powerful theorem gives us the nuke we were waiting for! * Insert devious laughter *
SD-3,4,5 theorem(For all nonnegative
tho).To verify if a symmetric homogeneous 3 variable polynomial
of degree is nonnegative for nonnegative , we just need to verify the cases:(i)
(ii)
Told you we'd get to nice stuff
Proof:
Every polynomial function in 3 variables of degree
can be written as 
. Fixing we maximise or minimise to find extrema of function. And extrema are achieved when two variables are equal or one is zero, and we're done!Based on similar logic, just that you need to focus on the Tej's theorem reasoning,to find this:
Let
be sides of a possibly degenerate triangle instead. Then we need to verify this just for:(i)
![$f(x,1,1) \geq 0 \forall x \in [0,2]$](http://latex.artofproblemsolving.com/3/9/e/39eaeeae6b9500155245b4c1cf1ba844777a2ea4.png)
(ii)
We also have a non-homogenous variant!
Non-homogenous variant: Let
satisfy all criteria that it did earlier, except homogeneity. We can still check for just the cases:(i)
(ii)
Some crucial facts that help you do manipulations as per your need:
(i)If the degree is
, you can make it a polynomial in instead and check that 
only from the second part of Tej's theorem!(ii) We used the fact that
has 
as functions. But they can be ANY functions, not just polynomials!Edit: Hey guys, back with part of what I promised!
SD-6 Theorem: Let
be degree 6 instead (and follows all other criteria as mentioned earlier). Then, it can be written as![\[ f(a,b,c) = Aw^6 + B(u,v^2)w^3 + C(u,v^2) \]](http://latex.artofproblemsolving.com/c/9/9/c992e31cbd796e035d9da84c0f2f439bc9688876.png)
After noting that 
is a constant, see that there are two cases:(i)
: Just check if and . The same logic as used in Tej's theorem.(ii) Um..... this is the bad part. For
, we need to prove![\[ f(a,b,c) \geq A\left(w^3 + Ku^3 + Luv^2 + M\dfrac{v^4}{u}\right)^2 \]](http://latex.artofproblemsolving.com/8/c/d/8cd30bcc953376100064a7fa43022afad049259b.png)
So yeah, we need to find 
ourselves. The point of doing this is to make 
to apply (i) now.Adding problems soon!
Dec 27:
Edit: Ok maybe not so soon
April 1:
Edit: Can we act like I never said this?
This post has been edited 14 times. Last edited by MathPassionForever, Apr 1, 2020, 12:54 PM
September 2019