 Posers and Puzzles

1. 25 Oct '07 22:272 edits
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".
2. 26 Oct '07 09:15
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".
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 >= 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.
3. 26 Oct '07 12:02
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.
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)
4. 26 Oct '07 12:14
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)
That wasn't the original inequality, though. And sqrt(ab + cd) < sqrt(ab) + sqrt(cd)
5. 26 Oct '07 12:16
Originally posted by mtthw
That wasn't the original inequality, though. And sqrt(ab + cd) < sqrt(ab) + sqrt(cd)
Yeah. I just noticed it now. I really need to learn how to read.
6. 26 Oct '07 12:24
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.
7. 26 Oct '07 18:33
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".
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.
8. 26 Oct '07 19:54
Originally posted by Kepler
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.
No it isn't.

0.5 >= 0.488...

The proof didn't assume anything other than positive numbers.
9. 26 Oct '07 20:08
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.
Well done!
😀
10. 27 Oct '07 07:52
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! 😀
11. 27 Oct '07 11:15
Originally posted by wolfgang59

"(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! 😀
The logic is fine. You can see that if you follow the steps in the demonstartion in the opposite direction.

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.
12. 27 Oct '07 13:09
The logic is fine. You can see that if you follow the steps in the demonstartion in the opposite direction.

Another way to see that is by noticing that I proved that the result that was intended was [b]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 w ...[text shortened]... y but english is not my first language and sometimes I may not express myself in the best way.[/b]
"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!
13. 27 Oct '07 15:31
The logic is fine. You can see that if you follow the steps in the demonstartion in the opposite direction.

Another way to see that is by noticing that I proved that the result that was intended was [b]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 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 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
14. 27 Oct '07 16:432 edits
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!
15. 27 Oct '07 16:44