View Full Version : Math Problem, Help needed

ibemadskillzz

09-11-2004, 11:56 AM

I was looking through my Calculus book from long time ago, and I came across a problem that I could not solve. It said, not to substitute numbers to solve the problem. It wants us to show proofs like geometry. The problem is: Prove that- Square root of AB is less than or equal to 1/2(a+b).

Sorry I do not have the square root button, but the AB is inside the square root sign. And 1/2- is One-half as in fraction. I also do not have to Less than or equal to sign to type.

thanks for all your help.

ibemadskillzz

09-12-2004, 05:43 AM

It's amazing how no one can solve this problem. I expected more out of you guys. Are there any Math teachers out there? help

aahala

09-12-2004, 07:45 AM

I don't know if this fits your requirements, but here's a proof involving algebra and geometry:

"Prove that- Square root of AB is less than or equal to 1/2(a+b). "

If we make this an equation and square both sides we get:

AB= 1/4(A2 +2AB + B2)

So:

4AB=A2 +2AB + B2

2AB=A2 + B2

The terms of the equation can be considered areas of rectangles, such as A2 is a square of sides A etc.

To prove the original claim, one can prove it's impossible to fit the areas represented on the right into the area on the left and have any "space" leftover.

If A=B, then it's obvious there's nothing left, as the equation becomes some form of: 2A2= A2 +A2 or 2B2 = B2 + B2.

We can let A or B be the larger number, it doesn't matter. Let's let B be the larger and represent it as A + X.

So we can stack two rectangles with dimensions A and B on top of one another, resulting in a rectangle B as the base and 2A in height for the left of the equation.

Take out the b2 rectangle from this. What's left is a rectangle(if there's anything left at all) a rectange height of A- X and base A + X.

Can we get A2 out of such a rectangle? No.

(A - X)(A +X)=A2

A2 - X2=A2 (This is not possible)

We have proven the left side can not be greater than the right side of the original equation, which was our objective.

ibemadskillzz

09-12-2004, 03:19 PM

thanks for your help. I kind of understand it, but I was looking for a all algebraic solution. Can someone prove it without using geometry shapes and formulas ?

thanks

aahala

09-12-2004, 04:21 PM

2AB=<A2 + B2

A+X=B

2A(A+X)=<A2 +(A+X)2

2A2 +2AX=<A2 +A2 +2AX +X2

2AX=<2AX +X2

0=<X2

Feņa14

09-14-2004, 09:07 AM

Wow that is crazy :shock:

Is there smoke coming out of your ears? :lol:

SirGus

09-20-2004, 03:03 PM

I was looking through my Calculus book from long time ago, and I came across a problem that I could not solve. It said, not to substitute numbers to solve the problem. It wants us to show proofs like geometry. The problem is: Prove that- Square root of AB is less than or equal to 1/2(a+b).

Sorry I do not have the square root button, but the AB is inside the square root sign. And 1/2- is One-half as in fraction. I also do not have to Less than or equal to sign to type.

thanks for all your help.

What's AB?a?b?

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