Olympiad inequality question

[Levido]

Hey, thanks for viewing my new thread where I get stuck on olympiad questions. This particular method isn’t mine but I understand it up to this point.

the question:


The steps so I understand:


Next step is applying AM-GM/Cauchy-Schwarz inequality (cyclic, so grouping x^2y +1/y 3 times) but I only have a surface understanding of what am-gm is and can’t see how it groups and is applied. What am I missing?

Here is a link to an AoPS thread about it:

Math Message Boards FAQ & Community Help | AoPS

artofproblemsolving.com

and here are the solutions:

https://media.hotnews.ro/media_server1/document-2014-05-6-17176147-0-subiecte-balcaniada-matematica-2014.pdf

(q1)
Im interested in the alternative method

Cheers for reading ?

 

[Cubist]

I think you want to see why the following is true:-
[math] x^2y + \frac{1}{y} \ge 2x [/math]
Multiply through by y (we know that y≥0 so the inequality stays pointing the same way)

[math] x^2y^2 + 1 \ge 2xy [/math]
[math] x^2y^2 - 2xy + 1 \ge 0 [/math]
Can you factor the LHS?

 

[Levido]

Can you factor the LHS?

Of course! What a brilliant idea. (xy-1)². I still can’t see how AM-GM can be applied but this works great.

We can also spot that if (xy-1)² =0, xy =1 and so does yz and xz and that means that for equality x,y and z= 1

 

[lex]

Next step is applying AM-GM/Cauchy-Schwarz inequality (cyclic, so grouping x^2y +1/y 3 times) but I only have a surface understanding of what am-gm is and can’t see how it groups and is applied. What am I missing?

I think you want to see why the following is true:-
[math] x^2y + \frac{1}{y} \ge 2x [/math]

AM-GM says that [MATH]\frac{a+b}{2}≥\sqrt{ab}[/MATH]Just apply that three times!
[MATH]a=x^2y, b=\frac{1}{y}[/MATH]

 

[lex]

I hope that post solved it for you.
Just out of interest also:
[MATH]a + \frac{1}{a}≥2[/MATH] for all [MATH]a\in \mathbb{R^+} \hspace2ex [/MATH] (easy to see, and easily proven).
So: [MATH]xy + \frac{1}{xy}≥2[/MATH]so [MATH]x^2y+\frac{1}{y}≥2x \hspace2ex (x, y \in \mathbb{R^+}[/MATH])

 

[Levido]

I hope that post solved it for you.

Yes it does completely. For me there are some things you need to be shown being used once before you fully understand how it can be applied.

Just out of interest also:
…

Good spot! I like how this builds from something towards the question, it’s undeniably true at every step so it’s more intuitive than applying a formula.

We can also spot that if (xy-1)² =0, xy =1 and so does yz and xz and that means that for equality x,y and z= 1

With hindsight I’m not sure how rigorous this was of me to say. All I know is that if you use the AM-GM inequality then the only way for equality to hold is for all the terms=1

Now time for me to go study the proof of the AM-GM inequality

Thank you for your guidance Lex and Cubist

 

[lex]

We can also spot that if (xy-1)² =0, xy =1 and so does yz and xz and that means that for equality x,y and z= 1

[MATH]xy=yz=zx \Rightarrow \frac{1}{z}=\frac{1}{x}=\frac{1}{y} \hspace2ex[/MATH] (dividing by [MATH]xyz[/MATH])
We know from the original equation in the question* that:
[MATH]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\\ \therefore \frac{1}{x}=\frac{1}{y}=\frac{1}{z}=1[/MATH]

* (dividing the original[MATH] \hspace1ex xy+yz+zy=3xyz [/MATH] by [MATH]xyz[/MATH])

 

[Levido]

[MATH]xy=yz=zx \Rightarrow \frac{1}{z}=\frac{1}{x}=\frac{1}{y} \hspace2ex[/MATH] (dividing by [MATH]xyz[/MATH])

Surely from the question we only know that 1/x +1/y+1/z=3, not that they’re all equal as you said? If they’re all equal then it’s just one instance of the inequality. Are you deriving this from somewhere?

Sorry to prolong this, I like to understandas much as I can.

 

[lex]

Surely from the question we only know that 1/x +1/y+1/z=3, not that they’re all equal as you said?

This was referring to your questioning the situation when equality holds.
When equality holds, you showed that xy=yz=zx
and I just showed that this implies that x=y=z=1.

 

[Steven G]

I think you want to see why the following is true:-
[math] x^2y + \frac{1}{y} \ge 2x [/math]
Multiply through by y (we know that y≥0 so the inequality stays pointing the same way)

[math] x^2y^2 + 1 \ge 2xy [/math]
[math] x^2y^2 - 2xy + 1 \ge 0 [/math]
Can you factor the LHS?

Cubist, that really is a brilliant move! Do you just know it cold that [math] x^2y + \frac{1}{y} \ge 2x [/math]? If not, can you explain what made you consider it?
Some people really have the gift. I just have a little of it and it comes very infrequently.

 

[pka]

Cubist, that really is a brilliant move! Do you just know it cold that [math] x^2y + \frac{1}{y} \ge 2x [/math]? If not, can you explain what made you consider it?

It is well known that if \(a>0\) then \(a+\dfrac{1}{a}\ge 2\) To see that square \(\sqrt{a}-\dfrac{1}{\sqrt{a}}\).
Because both \(x~\&~y\) are positive \(xy+\dfrac{1}{xy}\ge 2\). multiply through by \(x\).

 

[Steven G]

It is well known that if \(a>0\) then \(a+\dfrac{1}{a}\ge 2\) To see that square \(\sqrt{a}-\dfrac{1}{\sqrt{a}}\).
Because both \(x~\&~y\) are positive \(xy+\dfrac{1}{xy}\ge 2\). multiply through by \(x\).

Where else does it come up to make it well known? I too would like to have this as a tool. Thanks!

 

[pka]

Where else does it come up to make it well known? I too would like to have this as a tool.

Every mathematics library should have a copy of AN INTRODUCTION TO INEQUALITIES by Beckenbach & Bellman. I first encountered it there. They use it a good it there.

 

[Steven G]

Check out some of these prices for that book at

 

[Steven G]

I found a pdf version of the book here

 

[Cubist]

Cubist, that really is a brilliant move! Do you just know it cold that [math] x^2y + \frac{1}{y} \ge 2x [/math]? If not, can you explain what made you consider it?
Some people really have the gift. I just have a little of it and it comes very infrequently.

You have the gift of being able to teach in a classroom, and one-to-one (like some other helpers here)! I don't think that I could do that. I don't think on my feet fast enough to answer random questions on the spur of the moment.

My own thought process was very simple. I wondered why Levido had mentioned "AM-GM" when this wasn't mentioned in the original question. I had not heard of "AM-GM" before so I was curious. Therefore, I clicked on the first link in post#1 and then I searched within the page for "AM-GM". Reply #22 on that page, from "aditya21", was very interesting (click here for a direct link to that post). I still didn't know what "AM-GM" was. But the large inequality had been split into three equivalent smaller ones. At that point I thought I'd be able to prove one (and thus all) of these simple inequalities. Elimination of the fraction seemed a good first step, and this led to my post#2.

Not so brilliant after all. But I'm glad that I've now learned about AM-GM from the rest of this thread.