inequality

[triclino]

prove : [math] a^2+ab+b^2\geq 0[/math] for all a,b real Nos

[D H]

Hint: You should be able to readily prove that this is the case when ab<0. The cases where ab=0 are even easier. That leaves ab>0.

 

Complete the square.

[triclino]

I don't get you

[shyvera]

Treating the left-hand expression as a quadratic in a, verify that the determinant is never positive.

[triclino]

Treating the left-hand expression as a quadratic in a, verify that the determinant is never positive.

 

If we treat the L.H.S as quadratic we have:

 

[math]a = \frac{-b+\sqrt{b^2-4b^2}}{2}= \frac{-b+\sqrt{3}bi}{2}[/math]

 

.............................or.......................................................

 

[math]a = \frac{-b-\sqrt{b^2-4b^2}}{2}= \frac{-b-\sqrt{3}bi}{2}[/math]

 

Which makes, a ,a complex No

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Hint: You should be able to readily prove that this is the case when ab<0. The cases where ab=0 are even easier. That leaves ab>0.

 

Complete the square.

 

Yes, i completed the square and it works.

 

But what is all this about ab??

[D H]

If you complete the square you should get

 

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

 

 

Yet another way to prove this is to find a 2x2 matrix A such that

 

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

\bmatrix a & b \endbmatrix

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

 

There is only one such matrix, and it is positive definite.

[triclino]

If you complete the square you should get

 

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

 

.

 

The method that i know from high school in completing the square is the following :

 

[math]a^2+ab+b^2=a^2+2a(\frac{b}{2})+\frac{b^2}{4}-\frac{b^2}{4}+b^2[/math] which is equal to:

 

[math] (a+\frac{b}{2})^2 +\frac{3b^2}{4}[/math]. Which is definitely greater than or equal to zero for all values of a and b.

 

On the other hand to get : [math] (a+b)^2\geq ab[/math] you must assume [math]a^2+ab+b^2\geq 0[/math] ,which is what you want to prove

[D H]

Oh, please.

 

[math]a^2+ab+b^2 = a^2 + 2ab + b^2 - ab = (a+b)^2 - ab[/math]

[triclino]

Oh, please.

 

[math]a^2+ab+b^2 = a^2 + 2ab + b^2 - ab = (a+b)^2 - ab[/math]

 

If ab>0 => -ab<0 .But[math] (a+b)^2 >0[/math] .How can you add these two inequalities to get :

 

[math](a+b)^2 -ab>0[/math] and consequently [math](a+b)^2>ab[/math] ??

[D H]

[math](a+b)^2 -ab>0[/math] and consequently [math](a+b)^2>ab[/math] ??

Exactly.

 

Adding the same quantity to both sides of an inequality does not change the inequality.

[triclino]

Exactly.

 

Adding the same quantity to both sides of an inequality does not change the inequality.

 

 

Please, write down the inequality and the quantity that you add to both sides of the inequality

[D H]

No, for two reasons.

 

1. This looks too much like homework. You need to figure this out on your own.

2. You already did it in post #9.

[triclino]

No, for two reasons.

 

1. This looks too much like homework. You need to figure this out on your own.

2. You already did it in post #9.

 

Now i can see you have problems with inequalities. In post # 9 i showed exactly the opposite.

 

I showed that if ab>0 then you cannot prove [math](a+b)^2\geq ab[/math]

[D H]

Baloney.

 

Let's start with [math]a^2+ab+b^2 \ge 0[/math]. Adding 0 to both sides of the inequality does not change the inequality, so [math]a^2+ab+b^2 +ab -ab \ge 0 + 0[/math]. (Note that I write 0 in the form [math]ab-ab[/math] on the left hand side.) The left hand side is [math](a+b)^2 - ab[/math], and thus the original inequality is equivalent to [math](a+b)^2 - ab \ge 0[/math]. Adding [math]ab[/math] to both sides results in [math](a+b)^2 \ge ab[/math].

 

 

There is a simpler way to obtain the same result. Start again with [math]a^2+ab+b^2 \ge 0[/math]. Adding [math]ab[/math] to both sides yields [math]a^2+ab+b^2 + ab \ge ab[/math], or [math](a+b)^2 \ge ab[/math].

Edited May 28, 2010 by D H

[shyvera]

Now i can see you have problems with inequalities. In post # 9 i showed exactly the opposite.

 

I showed that if ab>0 then you cannot prove [math](a+b)^2\geq ab[/math]

 

You did no such thing. You were merely stuck somewhere, and being stuck doesn’t prove or disprove anything. You are the one having problems with proofs.

 

We also seem to be losing the plot a bit. We’re supposed to prove that [math]a^2+ab+b^2\geqslant0[/math] for all real a and b, not assume it’s true and use it to prove something else. So let’s go back a little in the thread.

 

In post #4, I suggested attacking the problem by the determinant method. Your reply clearly showed that you didn’t know what I meant, so let me explain. Consider the following expression:

 

[math]x^2+px+q[/math]

This is a quadratic in x. The determinant of the quadratic is the quantity [math]p^2-4q.[/math] It can be shown that [math]x^2+px+q\geqslant0\ \mbox{for all}\ x\in\mathbb R[/math] [math]\Leftrightarrow[/math] [math]p^2-4q\leqslant0.[/math] Well, let’s show it, shall we?

 

[math](\Leftarrow)[/math] Suppose [math]p^2-4q\leqslant0.[/math] Then a little rearrangement gives [math]q-\frac{p^2}4\geqslant0.[/math] Also [math]\left(x+\frac p2\right)^2\geqslant0[/math] for all real x. Adding the two inequalities gives [math]\left(x+\frac p2\right)^2+q-\frac{p^2}4\geqslant0[/math] and the left-hand side is just [math]x^2+px+q.[/math]

 

[math](\Rightarrow)[/math] Now suppose [math]x^2+px+q\geqslant0[/math] is true for all real x. In particular, it will be true for [math]x=-\frac p2.[/math] Substituting this into the inequality and simplifying should give [math]p^2-4q\leqslant0[/math] as required.

 

Now we return to your original problem. Treat the expression [math]a^2+ab+b^2[/math] as a quadratic in a. (Or a quadratic in b, if you like.) Now calculate the determinant. What do you find?

[triclino]

In post #7 i proved the inequality in concern .Read it if you like.

 

The hole proof is just a line .

 

I would like to comment on your proof ,but as usually the moderators

will intervene and give me valuable advice on how to behave e.t.c

 

Now,

on your comment that i have problems with proofs i can say the following:

 

To really say whether a proof is correct or not you must analyze the proof by writing a formal issue of it.

 

Then i will except you as capable of criticizing my proofs.

 

Can you write a formal proof of any high school theorem?

[D H]

I would like to comment on your proof ,but as usually the moderators will intervene and give me valuable advice on how to behave e.t.c

It would be good if you paid attention to that valuable advice.

 

You do not have to assume [math]a^2+ab+b^2\ge 0[/math] to prove [math](a+b)^2\ge ab[/math]. These two statements say exactly the same thing. This means of course that if you do prove the former you have automatically proven the latter. However, if you can prove the latter you have automatically proven the former. I thought you might have an easier time proving the latter statement.

[triclino]

It would be good if you paid attention to that valuable advice.

 

You do not have to assume [math]a^2+ab+b^2\ge 0[/math] to prove [math](a+b)^2\ge ab[/math]. These two statements say exactly the same thing. This means of course that if you do prove the former you have automatically proven the latter. However, if you can prove the latter you have automatically proven the former. I thought you might have an easier time proving the latter statement.

 

I suggest you go and read (perhaps in a book of logic) what is the double implication proof .Check the following example:

 

0x = 0 <=> 0x+x = 0+x <=> x(0+1) = 0 +x <=> x= 0+x <=> x=x

 

We want to prove 0x = 0 (for all real ,x) ,so by double implication we arrive at a well known fact x=x ,then we can except 0x=0 as true.

 

So to go from [math]a^2+ab+b^2\geq 0[/math] to [math] (a+b)^2\geq ab[/math] you do the following:

 

[math]a^2+ab+b^2\geq 0\Longleftrightarrow a^2+ab+b^2+ab\geq ab\Longleftrightarrow (a+b)^2\geq ab[/math]

Is [math](a+b)^2\geq ab[/math] a well known fact??

[Cap'n Refsmmat]

Perhaps your dedication to learning mathematics should be redirected to mastering your English?

 

D H is not proving [math]a^2+ab+b^2\geq 0[/math]. He is stating that it is equivalent to [math](a+b)^2\geq ab[/math]. If you prove the latter, you have proven the former. Try proving the latter.

[triclino]

Perhaps your dedication to learning mathematics should be redirected to mastering your English?

 

 

And you should learn to read the whole thread before you get involved in a discussion and come into worthless and impolite conclusions

 

 

D H is not proving [math]a^2+ab+b^2\geq 0[/math]. He is stating that it is equivalent to [math](a+b)^2\geq ab[/math]. If you prove the latter, you have proven the former. Try proving the latter.

 

If you prove the latter i will stop posting in this forum.

 

The latter is unprovable (unless you do the same mistake with D.H to use [math]a^2+ab+b^2\geq 0[/math] in your proof)

[D H]

Is [math](a+b)^2\geq ab[/math] a well known fact??

I am not going to spoon feed you the answers to your homework-like questions, triclino. That won't help you learn. I am instead trying to help you solve the problem mostly on your own with a little nudge here and there.

 

 

Suppose I told you that the eigenvalues of the matrix

 

[math]\bmatrix 1 & 1/2 \\ 1/2 & 1 \endbmatrix[/math]

 

are 1/2 and 3/2 and therefore [math]a^2+ab+b^2 \ge 0[/math] because

 

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

\bmatrix a & b \endbmatrix

\bmatrix 1 & 1/2 \\ 1/2 & 1 \endbmatrix \bmatrix a \\ b \endbmatrix[/math]

 

Did that help you? (No, it didn't.)

 

 

Spoon-feeding someone the answers is generally far easier than trying to help that person learn.

Edited May 29, 2010 by D H
LaTeX

[triclino]

D.H, why you throw your problems on me .I think this is unfair.

 

Besides i accept advice and teaching from anybody as long as it is correct.

[Cap'n Refsmmat]

And you should learn to read the whole thread before you get involved in a discussion and come into worthless and impolite conclusions

Believe me, I've been following this discussion from the beginning.

 

The latter is unprovable (unless you do the same mistake with D.H to use [math]a^2+ab+b^2\geq 0[/math] in your proof)

D H is trying to get you to work this out for yourself. Clearly you don't want to. I have done the proof myself and it's fairly trivial. Here's the first few steps:

 

[math](a+b)^2 =a^2 + 2ab + b^2[/math]

 

So we're asserting that

 

[math]a^2 + 2ab + b^2 \geq ab[/math]

 

Subtracting,

 

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

 

And I think you'll find that easy to prove.

 

Go on, try it, show us where you get.

 

edit: I now realize that what doing is probably far different than D H's method for proving the inequality. Oh well; there's many ways to get to the same solution.

Edited May 29, 2010 by Cap'n Refsmmat

[triclino]

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

[Cap'n Refsmmat]

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.