25 Oct '07 22:27>2 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".
"sqrt" is square root, ">=" means "is bigger than or equal to".
Originally posted by David113I'm sure there is a quicker proof - this is just off the top of my head .....
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".
Originally posted by wolfgang59A quicker proof:
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.
Originally posted by adam warlockThat wasn't the original inequality, though. And sqrt(ab + cd) < sqrt(ab) + sqrt(cd)
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)
Originally posted by David113I 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.
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".
Originally posted by wolfgang59Well done!
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.
Originally posted by adam warlockYour math is good but not your logic.
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.
Originally posted by wolfgang59The logic is fine. You can see that if you follow the steps in the demonstartion in the opposite direction.
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! 😀
Originally posted by adam warlock"You can see that if you follow the steps in the demonstartion in the opposite direction"
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]
Originally posted by adam warlockI agree that your logic is fine 🙂
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]
Originally posted by wolfgang59Have you read the second part of my reply?
"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!
Originally posted by TheMaster37Yes 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.
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