testing

testing

Hi guys, I'm new. I'm somewhat limited on my mathematical terms, so please go easy on me(I referred annular circles(or whatever it is called) as donut circles before actually knowing what is it called).

Eh, I found this somewhere, tried it, but it got irritating after a while(2 weeks, that is). The equation is obvious, but I am missing something.

Given that a>0, b>0 and a+b=1, prove that

(a+(1/a))^2 + (b+(1/b)^2 is greater or equal than 25/2

From the conditions and equation given, one can pretty much see that the equation is correct. But I tried proving it through algebriac means(Calculation, also known as 'hammering' the problem)... and umm... well... x_x

Tried replacing b with 1-a yet?

>>3

Yes. It's the first thing I did. 2 weeks. I can't miss that.

I even tried arranging them into a+b forms for each part area, in all but rapid failure.

No ideas, anyone?

Or nobody is trying/people did not notice/people ignored this thread? ._.

Wait? can't you do like
f(x)=(a + 1/a)^2 + (a-1+1/(a - 1))^2

f'(x)=-2/a²(a+1/a)-2/(a-1)²(a-1(a-1+1/(a - 1))

(-2/a²)*(a+1/a)-(2/(a-1)²)*(a-1+1/(a-1))=0   ->  a=0.5

Yeah and if you look at the graph of the function (f(x)=(a + 1/a)^2 + (a-1+1/(a - 1))^2) you can see that 0.5 is a minimum, therefore no other value of a can make the y-value drop lower than 25/2. Problem solved :D Hope you understood me, english maths is not my favourite.

I've got a feeling you're not supposed to use calculus.

>>6

Eh, maybe. Because I still have a little flaw in understanding the solution... I have an idea of what you're doing, but I'm a little fuzzy in the morningtime.

What is the form of the graph for the function? I just can't plainly imagine it. How does a quadic over quadic graph look like(Straight line? Lollerskates)? 0_o

Unless you are plotting points(middle of early morning, not doing it)... =_=ZzZ...

Try showing it in algebraic notations after you explain a bit further on your solution. Prove it by using both a and b in the inequality.

:( But it's the only way i know, except for just looking at it and just seeing that anything different from .5 makes it greater than 25/2 :P
You could draw it too, that's not calculus is it?^^

well it got 2 poles, one at 0 and one at 1 to +infinity , and between them is the minimum at exactly 0.5 , left and right of the poles the graph just looks like some missshaped parabola :D
hope you can post links on here (http://img115.imageshack.us/img115/9877/123cv2.gif)
This is what it looks like. And as neither b nor a can get lower or higher than 0 and 1. So i was just calculating the extreme values of the function. And as it has to be between 0 and 1 , i chose 0.5 of the three possibilities.

uhm, you could just replace all the a's with b's and it would still be the same :)

with f'(x) i meant the derivative.

The extreme value of the function is a minimum located at 0.5 / 12.5 , thus no lower value than 12.5 is possible.

Hope you understand what I'm trying to say, english isn't my first language and i never did maths in it before :D

I tend to just imagine the graph than to write it in paper. I do the planning first in my mind before doing the work on paper. This is something I cannot do at 3 am now.

Even straight out 0.5 as values for both a and b could work, I guess.They will add up to a number greater than 12.5. In fact, I know that any decimal numbers that a and b will add into one will produce a greater number than 12.5. That, I am well aware.

What does one mean by "It's the only way you know"?

Have your tried it in algebriac calculation already?

I guess I already have a grip on your solution... but the aim of me looking about is to see if it can be solved by algebriac means. Just to see a clearer form in which one can explain to an even slightly mathematically inclined person so he/she could understand.

Well, everything i do is to calculate the derivative, to prove that 0.5/12.5 is the minimum of the function. That is all by using algebraic rules.
(a+(1/a))^2 + (b+(1/b)^2 >=12.5
a+b=1
b=a-1
Now replace the b with (a-1) -> (a + 1/a)^2 + (a-1+1/(a - 1))^2
This is the function i'm using: f(x)=(a + 1/a)^2 + (a-1+1/(a - 1))^2

Now i calculate the derivative:
f'(x)= -2/a²(a+1/a)-2/(a-1)²(a-1(a-1+1/(a - 1))

And then I calculate the extreme value:

-2/a²(a+1/a)-2/(a-1)²(a-1(a-1+1/(a - 1)) = 0

a=0.5

f(0.5)=12.5

The minimum of the function is 0.5/12.5 , so it's impossible that you get any y value lower than 12.5.

*scratchscratch*

Umm... I do not know how to convey what I want to you...

Can you do it without using calculus, then? Just plain old moving about unknowns? Never mind if that can't be done. I just want a confirmation that one is screwed if one does that.

Your solution is already plausible. I am sure of it already. ._.

I don't think it's possible with just moving the unknowns, you'd get an equation with a^6...And i have no idea how to solve that one.
The only way i know is doing it by using calculus. And I think it's the only way to really prove it.

Fuck's sake.
Minimum of (a+1/a) is 2.5, when a = 1/2 (for a > 0).
a+b=1, so the minima of (a+1/a)+(b+1/b) coincide at a=b=1/2.
What more do you want?

>>15

Proof of the statement beginning "Minimum...", i.e. the actual crux of the problem.

But you can only calculate the minimum using calculus. :S

I don't get it, if not calculus than what do you want?

>>11
All real numbers between 0 and 1 have a decimal expansion, and thus, are decimal numbers.  "Feeling" that something is right isn't the same as a rigorous proof.

Furthermore, calculus is a subset of algebra.

>>15
Phail.  The minimum is 2 at a = 1.

>>1
The arithmetic mean - root mean square inequality states that for any x,y >0,
[(x² + y²)/2]^1/2 >= [x+y]/2, with equality iff x = y.

Thus,
(a+1/a)² + (b+1/b)² >= 2[(a+1/a+b+1/b)/2]² = (1/2)[1 + 1/a + 1/b]² = (1/2)[1 + (b+a)/(ab)]² = (1/2)[1 + 1/(ab)]²

ab = a(1-a) = a - a² = 1/4 - (a - 1/2)² <= 1/4

(1/2)[1 + 1/(ab)]² >= (1/2)[1 + 1/(1/4)]² = (1/2)[1 + 4]² = (1/2)5² = 25/2

>>19
Oops, I was referring to >>2, not >>1

For more information, see
http://en.wikipedia.org/wiki/Generalized_mean
(with p = 1,2 for AM, RMS respectively)

isn't that exactly what i said :S

>>19
(1+1/1)^2+(0+1/0)2 = 2?
What the fuck are you taking?
I mean, seriously, how could you even think that that is the case, especially when you have damn well proved it below.

>>22

Can you read?
I'm not >>19 btw

>>15 claimed (a + 1/a) ( and hence (a + 1/a)^2 ) was minimised at a=1/2 with (a + 1/a) = 2.5

This is obviously untrue as a = 1 gives (a + 1/a) = 2 < 2.5

But a can't equal 1, it's out of the given limits of a+b=1 and a and b being greater than 0