DEMONSTRATION HELP
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Dr.Peterson
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These are well-known inequalities, called the HM-GM-AM-QM inequalities; it appears that you are expected to prove them yourself. Where do you need help? Please show your work so we can see how far along you are.
You will probably want to take one inequality at a time.
You will probably want to take one inequality at a time.
pka
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If you are a student at institution with a mathematics library then look for
AN INTRODUCTION TO INEQUALITIES by Beckenbach & Bellman.
AN INTRODUCTION TO INEQUALITIES by Beckenbach & Bellman.
What is very weird here is that premise and conclusion are lumped into one line and that the premise itself is stated incompletely.
[MATH]\text {Given } 0 < x \le y, \text { prove}[/MATH]
[MATH]x \le \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \le \sqrt{xy} \le \dfrac{x + y}{2} \le \sqrt{\dfrac{x^2 + y^2}{2}} \le y.[/MATH]
Unlike Dr. Peterson, I would start with the case where 0 < x = y. Those are trivial.
Now move on to the more difficult case of 0 < x < y.
Then I might try
[MATH]x < \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \text { and } \sqrt{\dfrac{x^2 + y^2}{2}} < y.[/MATH]
A hint for the first of those.
[MATH]0 < x < y \implies \dfrac{1}{x} > \dfrac{1}{y} > 0 \implies \dfrac{1}{x} + \dfrac{1}{x} > \dfrac{1}{x} + \dfrac{1}{y} > 0.[/MATH]
Then I would do as Dr. Peterson advised: deal with each of the remaining inequalities separately.. For those, I have two suggested ideas, but warn that they may be dead ends because I have not tried them.
[MATH]\text {DEFINE: } z = y - x \implies y = x + z.[/MATH]
[MATH]0 < z < y \ \because \ 0 < x < y.[/MATH]
The other idea is that proofs by contradiction may be handy.
[MATH]\text {Given } 0 < x \le y, \text { prove}[/MATH]
[MATH]x \le \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \le \sqrt{xy} \le \dfrac{x + y}{2} \le \sqrt{\dfrac{x^2 + y^2}{2}} \le y.[/MATH]
Unlike Dr. Peterson, I would start with the case where 0 < x = y. Those are trivial.
Now move on to the more difficult case of 0 < x < y.
Then I might try
[MATH]x < \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \text { and } \sqrt{\dfrac{x^2 + y^2}{2}} < y.[/MATH]
A hint for the first of those.
[MATH]0 < x < y \implies \dfrac{1}{x} > \dfrac{1}{y} > 0 \implies \dfrac{1}{x} + \dfrac{1}{x} > \dfrac{1}{x} + \dfrac{1}{y} > 0.[/MATH]
Then I would do as Dr. Peterson advised: deal with each of the remaining inequalities separately.. For those, I have two suggested ideas, but warn that they may be dead ends because I have not tried them.
[MATH]\text {DEFINE: } z = y - x \implies y = x + z.[/MATH]
[MATH]0 < z < y \ \because \ 0 < x < y.[/MATH]
The other idea is that proofs by contradiction may be handy.
Last edited:
pka
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hoosie, the above post is completely useless. It may be correct or incorrect, how are we to tell?
You handwriting is absolutely unreadable. To form an x you make a backwards c next to a c.
Why do you think that someone who is struggling to understand the mathematics should further struggle to read your chicken scratches?
Your contributions are welcome as long as they are readable. This one is not.
If you are a serious about mathematics then you must learn to use LaTeX. If you do not, your posts will be for naught.
Ask yourself this question: "Would I bother to read a reply in a script that is hard to understand"?
D
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I do not think Hoosie's handwriting unintelligible. Hoosie's answers are more geared towards students than lot of others in this web-site. Hoosie provided part-solution to the problem and encouraged the student to solve the rest. Exactly what a good tutor should do. I welcome hand-written "help" from Hoosie (just don't provide "complete" solution. Leave something to discover for the student - like you did in your first response).
I applaud hoosie's use of a much more legible presentation, and I fully agree with Subhotosh Khan that it is a very rare case when we should give complete or almost complete answers before the OP has shown any work or even thought. I also agree that hoosie seems attuned to how beginning students think.
I hope, therefore, it will not discourage hoosie from future contributions to point out that this contribution is worthless.
His attempted proof that [MATH]\dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \le \sqrt{xy}[/MATH]
starts with
[MATH]\text {We know } \sqrt{xy} \le \dfrac{x + y}{2}. [/MATH]
We do not know that YET. It is one of the things that is to be proved. In fact, his very next "proof" is to demonstrate
[MATH]\sqrt{xy} \le \dfrac{x + y}{2}[/MATH],
and that proof relies on the proposition that
[MATH]\dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \le \sqrt{xy}.[/MATH]
The argument is completely circular. He has in fact proved only that
[MATH]\text {If } \sqrt{xy} \le \dfrac{x + y}{2} \text { is true, then } \sqrt{xy} \le \dfrac{x + y}{2} \text { is true.}[/MATH]
I hope, therefore, it will not discourage hoosie from future contributions to point out that this contribution is worthless.
His attempted proof that [MATH]\dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \le \sqrt{xy}[/MATH]
starts with
[MATH]\text {We know } \sqrt{xy} \le \dfrac{x + y}{2}. [/MATH]
We do not know that YET. It is one of the things that is to be proved. In fact, his very next "proof" is to demonstrate
[MATH]\sqrt{xy} \le \dfrac{x + y}{2}[/MATH],
and that proof relies on the proposition that
[MATH]\dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} \le \sqrt{xy}.[/MATH]
The argument is completely circular. He has in fact proved only that
[MATH]\text {If } \sqrt{xy} \le \dfrac{x + y}{2} \text { is true, then } \sqrt{xy} \le \dfrac{x + y}{2} \text { is true.}[/MATH]
Dr.Peterson
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I'll just make a couple comments on the matter of appearance.
First, Hoosie's handwritten "x" is standard in many parts of the world (though occasionally written too loosely); I have seen it often, particularly from foreign students in my face-to-face classes, and would never criticize it.
But second, I rarely bother to read Hoosie's answers, because they are far too large (whether handwritten or typed), and I don't care to read screen after screen of material just to check someone's work. Is this something that could be fixed by using the software differently?
First, Hoosie's handwritten "x" is standard in many parts of the world (though occasionally written too loosely); I have seen it often, particularly from foreign students in my face-to-face classes, and would never criticize it.
But second, I rarely bother to read Hoosie's answers, because they are far too large (whether handwritten or typed), and I don't care to read screen after screen of material just to check someone's work. Is this something that could be fixed by using the software differently?
pka
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Thank you for that reply. Clarity is so important to a student who is already struggling with understanding the material in the first place.
Hoosie
I am not sure that graphing constitutes a proof, but you are definitely on the right track.
[MATH]0 < x < y \implies \dfrac{1}{x} > \dfrac{1}{y} > 0.[/MATH] Very good.
The rest is simpler than you make it.
[MATH]\therefore \dfrac{1}{x} + \dfrac{1}{x} > \dfrac{1}{x} + \dfrac{1}{y} > \dfrac{1}{y} + \dfrac{1}{y} > 0 \implies[/MATH]
[MATH]\dfrac{2}{x} > \dfrac{x + y}{xy} > \dfrac{2}{y} > 0 \implies \dfrac{1}{x} > \dfrac{x + y}{2xy} > \dfrac{1}{y} > 0 \implies[/MATH]
[MATH]x < \dfrac{2xy}{x + y} < y \implies x < \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} < y.[/MATH]
I am not sure that graphing constitutes a proof, but you are definitely on the right track.
[MATH]0 < x < y \implies \dfrac{1}{x} > \dfrac{1}{y} > 0.[/MATH] Very good.
The rest is simpler than you make it.
[MATH]\therefore \dfrac{1}{x} + \dfrac{1}{x} > \dfrac{1}{x} + \dfrac{1}{y} > \dfrac{1}{y} + \dfrac{1}{y} > 0 \implies[/MATH]
[MATH]\dfrac{2}{x} > \dfrac{x + y}{xy} > \dfrac{2}{y} > 0 \implies \dfrac{1}{x} > \dfrac{x + y}{2xy} > \dfrac{1}{y} > 0 \implies[/MATH]
[MATH]x < \dfrac{2xy}{x + y} < y \implies x < \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{y}} < y.[/MATH]