inequality

[triclino]

No. I want to see you try it.

 

First try it for the case where a and b are both greater than 0. That should be trivial.

 

 

O.k teacher here we go:

 

case 1 : ab>0 ,then -ab<0 and since [math]0\leq a^2+b^2[/math] we have:

 

[math] -ab\leq a^2+b^2[/math]

 

case 2 : ab=0 ,then -ab=0 and since [math]0\leq a^2+b^2[/math] we have:

 

[math] -ab\leq a^2+b^2[/math]

 

 

case 3 : ab<0 ,then -ab>0. And here is the problem. Now we cannot say :

 

 

[math] -ab\leq a^2+b^2[/math].

 

Because if you have two Nos x>0 and y>0 you cannot say x>y

[Cap'n Refsmmat]

For case 3, let [math]|a| > |b|[/math].

 

[math]-ab < a^2[/math], correct? Both are positive, and [math]|b| < |a|[/math], so it must be true.

 

You can do it likewise for [math]|b| > |a|[/math], and [math]|b|=|a|[/math] is trivial.

 

This should let you complete the proof.

[triclino]

For case 3, let [math]|a| > |b|[/math].

 

[math]-ab < a^2[/math], correct? Both are positive, and [math]|b| < |a|[/math], so it must be true.

 

You can do it likewise for [math]|b| > |a|[/math], and [math]|b|=|a|[/math] is trivial.

 

This should let you complete the proof.

 

 

 

.................................WRONG.....................................................

 

This proof is completely wrong .Go over your proof again .Ask a teacher a Doctor ,anyone . This proof is wrong

[Cap'n Refsmmat]

It would be helpful if you were to explain why it is wrong. Just shouting "IT'S WRONG!" is obnoxious and unhelpful.

 

Anyway, if [math]|a| > |b|[/math] and [math]-ab>0[/math], a or b must be opposite in sign. If we take the expression I was showing,

 

[math]-ab < a^2[/math]

 

and divide through by a,

 

[math]-b< a[/math]

 

we can see this is equivalent to my condition, [math]|a| > |b|[/math].

 

And since [math]a^2 < -ab[/math], [math]a^2 + b^2 \geq -ab[math], since [math]b^2\geq 0[/math].

Edited May 29, 2010 by Cap'n Refsmmat

[D H]

Capn's proof is correct, triclino.

 

Your attitude is a huge hindrance to your learning.

 

Since you insist on being spoon-fed, here are some big servings.

 

 

Method 1

By inspection,

 

[math]a^2+ab+b^2 = 3\left(\frac{a+b}2\right)^2 + \left(\frac{a-b}2\right)^2[/math]

 

The two terms on the right are non-negative for all real a and b. The sum of two non-negative values is non-negative, and therefore a^2+ab+b^2 is non-negative, or in other words

 

[math]a^2+ab+b^2 \ge 0[/math]

 

 

 

Method 2

Viewing a^2+ab+b^2 as a quadratic in a, the zeros are given by

 

[math]a = \frac{-b \pm \sqrt{-3b^2}}2[/math]

 

When b=0 the two solutions are degenerate and are zero. In this case, the expression a^2+ab+b^2 degenerates to a^2, and this is non-negative for all real a. When b≠0, the two solutions are imaginary. Thus a^2+ab+b^2 has the same sign for all real a. At a=0, a^2+ab+b^2 degnerates b^2, which is positive for all non-zero real b. Thus

 

[math]a^2+ab+b^2 \ge 0[/math]

 

 

 

Method 3

By inspection, [math]a^2+ab+b^2\ge 0[/math] is equivalent to [math]a^2+b^2\ge -ab[/math]. This is trivially true when ab≥0. This leaves the case ab<0 to be investigated. Define

 

[math]

\aligned

\alpha &\equiv \max(|a|, |b|) \\

\beta &\equiv \min(|a|, |b|)

\endaligned

[/math]

 

In terms of these new variables, [math]a^2+b^2=\alpha^2+\beta^2[/math] and in the case where [math]ab<0[/math], [math]-ab = \alpha\beta[/math]. Note that by definition, [math]\alpha\ge\beta[/math]. Multiplying both sides by [math]\alpha[/math] does not change the inequality since [math]\alpha>0[/math]. Thus [math]\alpha^2\ge\alpha\beta[/math]. Since [math]\beta \ge 0[/math] by definition, [math]\alpha^2+\beta^2\ge \alpha^2[/math]. As greater than is transitive, so [math]\alpha^2+\beta^2\ge \alpha\beta[/math]. In terms of the original variables,

 

[math]a^2+b^2\ge -ab[/math]

 

when ab<0. Combining the cases ab≥0 and ab>0 yields

 

[math]a^2+b^2\ge -ab[/math]

 

for all real a, b, and thus

 

[math]a^2+ab+b^2 \ge 0[/math]

 

 

 

Method 4

By inspection, [math]a^2+ab+b^2\ge 0[/math] is equivalent to [math](a+b)^2\ge ab[/math]. This is trivially true when ab≤0. This leaves the case ab>0 to be investigated. Define

 

[math]

\aligned

\alpha &\equiv \max(|a|, |b|) \\

\beta &\equiv \min(|a|, |b|)

\endaligned

[/math]

 

In terms of these new variables, [math](a+b)^2=(\alpha+\beta)^2[/math] and in the case where [math]ab>0[/math], [math]ab = \alpha\beta[/math]. Since [math]\alpha\ge\beta[/math] by definition, [math]\alpha^2+\beta^2>\alpha^2\ge\alpha\beta[/math] and thus

 

[math]a^2+ab+b^2 \ge 0[/math]

 

 

 

Method 5

Viewing a^2+ab+b^2 as a quadric in a and b,

 

[math]a^2 + ab + b^2 =

\bmatrix a & b \endbmatrix

\boldsymbol A \bmatrix a \\ b \endbmatrix [/math]

 

where

 

[math]\boldsymbol A = \bmatrix 1 & 1/2 \\ 1/2 & 1 \endbmatrix[/math]

 

The eigenvalues of the matrix A are 3/2 and 1/2 (verification left as an exercise to the reader). The matrix A is thus positive definite and thus serves as a metric on the space [math](a,b)\in\mathbb R^2[/math]. Thus

 

[math]a^2+ab+b^2 \ge 0[/math]

 

 

 

Method 6

Method 5 suggests expressing a and b in terms of the eigenvectors of the matrix A, which are [math]\bmatrix 1/\surd 2 & 1/\surd 2\endbmatrix^T[/math] and [math]\bmatrix 1/\surd 2 & -1/\surd 2\endbmatrix^T[/math]. By inspection,

 

[math]

\aligned

a &\equiv \frac{a+b}2 + \frac{a-b}2 \\

b &\equiv \frac{a+b}2 - \frac{a-b}2

\endaligned

[/math]

 

Thus

 

[math]

\aligned

a^2 &=\left(\frac{a+b}2\right)^2+

2\left(\frac{a+b}2\right)\left(\frac{a-b}2\right)+

\left(\frac{a-b}2\right)^2 \\

b^2 &=\left(\frac{a+b}2\right)^2-

2\left(\frac{a+b}2\right)\left(\frac{a-b}2\right)+

\left(\frac{a-b}2\right)^2 \\

ab &= \left(\frac{a+b}2\right)^2 - \left(\frac{a-b}2\right)^2

\endaligned

[/math]

 

and hence

 

[math]

a^2+ab+b^2 =3\left(\frac{a+b}2\right)^2 + \left(\frac{a-b}2\right)^2

[/math]

 

See method 1.

Edited May 29, 2010 by D H
I'm dyslexic (can't tell left from right)

[triclino]

For a start ,if i were to say that your method 2 is wrong what would you say??

[D H]

I would say "prove it".

[triclino]

If i were to prove that your proof is not correct we will have to formally analyze it

 

But can you formally analyze your own proof??

[D H]

Do you understand that a function f(x) that is quadratic in some variable x will necessarily have the same sign for all real x if the roots are not real-valued? This is very basic stuff, triclino.

[triclino]

So i take it that you cannot formally analyze your own proof??

[D H]

Just because you do not understand something does not mean it is wrong, triclino.

[Mr Skeptic]

For a start ,if i were to say that your method 2 is wrong what would you say??

 

I'd say you're making baseless accusations.

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Merged post follows:

Consecutive posts merged

No, No you curry on and finish the proof because i want stop posting in this forum

 

Well, he did finish the proof in several different manners. What do you have to say now?

[mooeypoo]

Thread moved to Homework Help, where it belongs. Triclino, you're very lucky our experts feed you the answers with a spoon; but I remind you that the proper action to do in cases where you get stuck on a question is to ask for help, rather than order people around.

 

Asking for help includes showing what you've already tried and failed, and why you think you need help. We aren't in the habit of feeding the answers back, or acting as online calculators. Please take that into account next time.

 

~moo

[triclino]

Thread moved to Homework Help, where it belongs. Triclino, you're very lucky our experts feed you the answers with a spoon; but I remind you that the proper action to do in cases where you get stuck on a question is to ask for help, rather than order people around.

 

Asking for help includes showing what you've already tried and failed, and why you think you need help. We aren't in the habit of feeding the answers back, or acting as online calculators. Please take that into account next time.

 

~moo

 

Mooeypoo if i was to ask your experts, for example, for a formal proof of a high school question they would loose their spoon .

 

There are a lot of things that your experts do not know.

 

If for example i was to ask your experts for a formal proof in Analysis probably i would wait centuries for an answer .

 

Boasting around is not a good thing.

 

I can even ask questions in Analysis (not formal ) that your experts wont know the answer.

 

So if you are so sure about your experts let us try .

 

I am not ordering anybody around.

[mooeypoo]

Mooeypoo if i was to ask your experts, for example, for a formal proof of a high school question they would loose their spoon .

 

There are a lot of things that your experts do not know.

 

If for example i was to ask your experts for a formal proof in Analysis probably i would wait centuries for an answer .

 

Boasting around is not a good thing.

 

I can even ask questions in Analysis (not formal ) that your experts wont know the answer.

 

So if you are so sure about your experts let us try .

 

I am not ordering anybody around.

"Solve!" is not very polite, triclino. And your attitude tends to be not very polite either. I can promise you that it takes the experts a lot longer to try and help you than it does to give you a straight up answer. And yet, they make an effort and try to see where the problem lies.

 

Instead of trying to fight them through it, or look for "AH HA!" moments you obviously miss, I would try and be a bit more courteous and a lot less short with the people who know more and spare their valuable time in helping me understand concepts I get stuck on.

 

No one owes you anything triclino, and people would elect to help you much more if you treat them a bit nicer. I don't know if you've noticed, but people *ARE* answering you. And you seem to be having trouble with the answers, looking for proof everywhere, even when things are trivial. That might be fine, but the fact *YOU* don't know the proof doesn't mean no one does, or that there is none.

 

Drop the attitude, be nicer, and you'll see how much better your experience (and others') is.

 

It will also be compatible with our rules and regulations, which demand a good attitude, and are non-negotiable.

 

~moo

 

By the way, if you still don't understand what I mean about attitude, read your reply again. These aren't "My" expert, and they're not "yours" either. Every time you challenged one of them, you came out flattened by your own statements. I recommend you be careful who you blame as unknowing.

And finally, if you don't like the people who post here, don't participate in debates here.

 

We will not accept condescending attitude for long, triclino.

[Mr Skeptic]

There are a lot of things that your experts do not know.

 

Of course, but one of the things they do know is that we're not supposed to spoon-feed answers. Remember also that there's a difference between not knowing an answer and not wanting to bother to tell you an answer (nor even give you hints). It really doesn't help any when you falsely pretend that a proof someone gave you is invalid. Whether that is due to incompetence or malice really doesn't matter: it makes you look bad. If you do not understand a proof there is no shame in saying so; pretending instead that it is invalid won't score you any points. I really wouldn't blame folks for totally ignoring you after the way you've treated them.

[Dave]

If i were to prove that your proof is not correct we will have to formally analyze it

 

But can you formally analyze your own proof??

 

Wow. I mean, I hate to break your bubble here but D H's post easily constitutes precisely what any sane mathematician would regard as a 'formal post'.

 

I don't have much to add to Capn's and mooey's posts here other than I'd advise the experts to give your posts a very wide berth from this point forwards. It seems to me like you're less interested in learning mathematics than constantly starting arguments over trivial proofs such as these.