# Prove the inequality

David113
Posers and Puzzles 25 Oct '07 22:27
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. wolfgang59
Mr. Wolf
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.
Baby Gauss
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)
Baby Gauss
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.
Baby Gauss
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. Kepler
Demon Duck
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!
ðŸ˜€
10. wolfgang59
Mr. Wolf
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.

ðŸ˜•

Rework your proof and it is much better than mine! ðŸ˜€
Baby Gauss
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.

ðŸ˜•

Rework your proof and it is much better than mine! ðŸ˜€
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. wolfgang59
Mr. Wolf
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. TheMaster37
Kupikupopo!
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 ðŸ™‚

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
Baby Gauss
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!
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?