- 26 Oct '07 09:15

I'm sure there is a quicker proof - this is just off the top of my head .....*Originally posted by David113***Prove: for positive numbers a, b, c, d, sqrt((a+d)(b+c)) >= sqrt(ab)+sqrt(cd).**

"sqrt" is square root, ">=" means "is bigger than or equal to".

First I need to prove x+y>=2sqrt(xy)

Lets assume x>=y so that we can write x+d=y where d>=0

Therefore d^2 >= 0

Therefore 4x^2 + 4xd + d^2 >= 4x^2 + 4xd

Factorising (2x + d)(2x + d) >= 4x(x+d)

Take roots (2x + d) >= 2sqrt(x{x+d})

substitute y for x+d .. x + y >= 2sqrt(xy) QED

Now back to original problem!!

We know (x+y) >= 2sqrt(xy) (I just proved it)

Let x = ac and y= bd

So we can write (ac + bd) >= 2sqrt(acbd)

Lets add ab + cd to both sides (just for fun!)

So ab + ac + bd + cd >= ab + cd + 2 sqrt(abcd)

Factor both sides

So (a+d)(b+c) >= {sqrt(ab) + sqrt(cd)}{sqrt(ab) + sqrt(cd)}

Take root of both sides

sqrt{(a+d)(b+c)} >= sqrt(ab) + sqrt(cd) Q.E.D.

Fingers crossed I have made no assumptions other than

that a, b, c, d are all positive. - 26 Oct '07 12:02

A quicker proof:*Originally posted by wolfgang59***I'm sure there is a quicker proof - this is just off the top of my head .....**

First I need to prove x+y>=2sqrt(xy)

Lets assume x>=y so that we can write x+d=y where d>=0

Therefore d^2 >= 0

Therefore 4x^2 + 4xd + d^2 >= 4x^2 + 4xd

Factorising (2x + d)(2x + d) >= 4x(x+d)

Take roots (2x + d) >= 2sqrt(x{x+d})

substitute y for x+d .. x + y > ...[text shortened]... .

Fingers crossed I have made no assumptions other than

that a, b, c, d are all positive.

Since all numbers are positive any product between them is positive too. So:

(a+d)(b+c)=ab+ac+db+dc > ab+cd Since ac,db > 0.

Now since sqrt is a monotonic function (I think that's how you sya it in english) we have sqrt((a+d)(b+c)) > sqrt(ab+cd) - 26 Oct '07 12:14

That wasn't the original inequality, though. And sqrt(ab + cd) < sqrt(ab) + sqrt(cd)*Originally posted by adam warlock***A quicker proof:**

Since all numbers are positive any product between them is positive too. So:

(a+d)(b+c)=ab+ac+db+dc > ab+cd Since ac,db > 0.

Now since sqrt is a monotonic function (I think that's how you sya it in english) we have sqrt((a+d)(b+c)) > sqrt(ab+cd) - 26 Oct '07 12:24The proof is now here:

sqrt((a+d)(b+c)) >= sqrt(ab)+sqrt(cd)

(a+d)(b+c) >= ab+cd+2sqrt(abcd)

ab+ac+db+dc >= ab+cd+2sqrt(abcd)

ac+db >= 2sqrt(abcd)

Let us denote ac=A and db=B. The former inequality is equivalent to

A+B >= 2sqrt(AB)

A+B-2sqrt(AB) >=0

(sqrt(A)-sqrt(B))^2 >= 0 Which is a valid statment so the initial assumption is right. - 26 Oct '07 18:33

I think you need to specify positive integers. For any positive numbers (the reals in effect) a = 0.1, b = 0.2, c = 0.3 and d = 0.4 is a counterexample.*Originally posted by David113***Prove: for positive numbers a, b, c, d, sqrt((a+d)(b+c)) >= sqrt(ab)+sqrt(cd).**

"sqrt" is square root, ">=" means "is bigger than or equal to". - 26 Oct '07 20:08

Well done!*Originally posted by wolfgang59***I'm sure there is a quicker proof - this is just off the top of my head .....**

First I need to prove x+y>=2sqrt(xy)

Lets assume x>=y so that we can write x+d=y where d>=0

Therefore d^2 >= 0

Therefore 4x^2 + 4xd + d^2 >= 4x^2 + 4xd

Factorising (2x + d)(2x + d) >= 4x(x+d)

Take roots (2x + d) >= 2sqrt(x{x+d})

substitute y for x+d .. x + y > ...[text shortened]... .

Fingers crossed I have made no assumptions other than

that a, b, c, d are all positive. - 27 Oct '07 07:52

Your math is good but not your logic.*Originally posted by adam warlock***The proof is now here:**

sqrt((a+d)(b+c)) >= sqrt(ab)+sqrt(cd)

(a+d)(b+c) >= ab+cd+2sqrt(abcd)

ab+ac+db+dc >= ab+cd+2sqrt(abcd)

ac+db >= 2sqrt(abcd)

Let us denote ac=A and db=B. The former inequality is equivalent to

A+B >= 2sqrt(AB)

A+B-2sqrt(AB) >=0

(sqrt(A)-sqrt(B))^2 >= 0 Which is a valid statment so the initial assumption is right.

"(sqrt(A)-sqrt(B))^2 >= 0 Which is a valid statment so the initial assumption is right."

Proving a valid statement from a conjecture does NOT infer the initial conjecture is correct!

eg I want to prove Socrates was an elephant.

1. Socrates was an elephant.

2. Elephants are mortal. ... "Which is a valid statement so the initial assumption is right"

3. Therefore Socrates was mortal.

Rework your proof and it is much better than mine! - 27 Oct '07 11:15

The logic is fine. You can see that if you follow the steps in the demonstartion in the opposite direction.*Originally posted by wolfgang59***Your math is good but not your logic.**

"(sqrt(A)-sqrt(B))^2 >= 0 Which is a valid statment so the initial assumption is right."

Proving a valid statement from a conjecture does NOT infer the initial conjecture is correct!

eg I want to prove Socrates was an elephant.

1. Socrates was an elephant.

2. Elephants are mortal. ... "Which is a valid st ...[text shortened]... herefore Socrates was mortal.

Rework your proof and it is much better than mine!

Another way to see that is by noticing that I proved that the result that was intended was**equivalent**to another statement:

sqrt((a+d)(b+c)) >= sqrt(ab)+sqrt(cd) < - > (SQRT(A)-SQRT(B))^2 >= 0 So both statements must be simultaneously wrong or simultaneously right. And since the second one is right so must be the first one.

In your example things don't work out cause the statements are not equivalent.

I'm sorry but english is not my first language and sometimes I may not express myself in the best way. - 27 Oct '07 13:09

"You can see that if you follow the steps in the demonstartion in the opposite direction"*Originally posted by adam warlock***The logic is fine. You can see that if you follow the steps in the demonstartion in the opposite direction.**to another statement:

Another way to see that is by noticing that I proved that the result that was intended was [b]equivalent

sqrt((a+d)(b+c)) >= sqrt(ab)+sqrt(cd) < - > (SQRT(A)-SQRT(B))^2 >= 0 So both statements must be simultaneously w ...[text shortened]... y but english is not my first language and sometimes I may not express myself in the best way.[/b]

I agree! So why not set them out in the correct direction? But if you are doing math in a second language I applaud you regardless! - 27 Oct '07 15:31

I agree that your logic is fine*Originally posted by adam warlock***The logic is fine. You can see that if you follow the steps in the demonstartion in the opposite direction.**to another statement:

Another way to see that is by noticing that I proved that the result that was intended was [b]equivalent

sqrt((a+d)(b+c)) >= sqrt(ab)+sqrt(cd) < - > (SQRT(A)-SQRT(B))^2 >= 0 So both statements must be simultaneously w ...[text shortened]... y but english is not my first language and sometimes I may not express myself in the best way.[/b]

Though What you gave us is what you usually work out on a side-paper.

To give the proof you would give the steps in opposite direction, starting with the obviously true statement.

Well, that's how my teachers told me to give proofs :p - 27 Oct '07 16:43 / 2 edits

Have you read the second part of my reply?*Originally posted by wolfgang59***"You can see that if you follow the steps in the demonstartion in the opposite direction"**

I agree! So why not set them out in the correct direction? But if you are doing math in a second language I applaud you regardless!

I have two equivalent prepositions; one of them is true so the other has no choice to be true too. So there is no right direction. Each way is true and the way you chose to expose your reasoning is a matter of personal taste. I rather be clear while expressing my reasoning so I saw no good in writing the steps in the opposite direction in which they came to me.

**But if you are doing math in a second language I applaud you regardless!**What does this means? - 27 Oct '07 16:44

Yes that's the way a mathematician would do it, but I'm physicist and normally we like to expose things how they came to us cause normally that's the best way to understand them.*Originally posted by TheMaster37***I agree that your logic is fine**

Though What you gave us is what you usually work out on a side-paper.

To give the proof you would give the steps in opposite direction, starting with the obviously true statement.

Well, that's how my teachers told me to give proofs :p