*** Problems

ashwil · 2006-03-07 01:32:57

***5

Ok, here goes, maybe a little clumsy and probably not 100% proof, but here goes:

for values of a,b,c greater than or equal to 1:-

If a, b & c all = 1, then (a + 1)^7 * (b+1)^7 * (c+1)^7 = 2,097,152 and 7^7* a^4 * b^4 * c^4 = 823,543
any increase in a,b or c can be viewed simply as a comparison between the functions (a+1)^7 and a^4. For all positive numbers greater than 1, this only increases the divergence. Therefore, it is only necessary to consider values of a,b,c between 0 & 1.

As a,b,c tend to zero, [(a+1)^7....] tends to an minimum value of 1, while [7^7.....] tends to zero. Again, as a,b,c increase above zero, the effect on the comparison is the same as for values above one. Namely, any increase in (a+1)^7 will be greater than the corresponding increase in a^4.

Obviously, negative values would reverse this, but for all positive values, the expression must be true, though I cannot actually see a point at which the expressions could ever be equal!

Jai Ganesh · 2006-03-07 04:33:47

ashwil, I shall post the solution to ***3 tomorrow. I was occupied with the other topics today. (Particularly, my post on quotes of mathematicians in 'Members only').

The solution to ***5, as you admitted, is not a 100% proof. You have considered certain values, and I think your reasoning is good, although it does not constitute a mathematical proof. I shall post the proof by the weekend.

ashwil · 2006-03-07 06:53:10

I'll settle for my reasoning being good. When you don't spend your time actually doing mathematical proofs, you do forget the notation, the methodology and the formulae, but reasoning powers can still get you a long way!

Ricky · 2006-03-07 07:24:15

***5

To simplify it, I'm going to represent a, b, and c by x, because you have to repeat this three times (once for each letter).

Thus, it becomes apparent we must show:

Where

If x < 1, then

Since all of the negative powers of x must be greater than or equal to 1.

The same reasoning goes for x > 2,

But I can't seem to get the numbers in between 1 and 2.

Edit:

How's this for an argument?

The graph

is continuous, above, and never intersects with 94 for all positive values x less than 2. Thus, it must always be greater there.

Last edited by Ricky (2006-03-07 07:36:19)

krassi_holmz · 2006-03-07 07:36:52

Logaritms?

krassi_holmz · 2006-03-07 07:47:38

Can someone do this:
7(ln(a + 1) + ln(b + 1) + ln(c + 1) - ln 7) ≥ 4ln abc

krassi_holmz · 2006-03-07 08:00:57

Well done, Ricky!
This was great idea!
Your function has local minimum between 1 and 2 at x=4/3.

krassi_holmz · 2006-03-07 08:21:51

***5
Something stinks here.
And where's the ricky's post?

krassi_holmz · 2006-03-07 08:23:24

Let

(ricky's function)

krassi_holmz · 2006-03-07 08:25:00

Here's a plot:

Last edited by krassi_holmz (2006-03-07 08:25:44)

krassi_holmz · 2006-03-07 08:28:51

We need to find the minimum for x>0 (actually we don't need the other side because a,b,c are positive).
How?
Calculus, of course!:

Last edited by krassi_holmz (2006-03-07 08:29:21)

krassi_holmz · 2006-03-07 08:40:39

The main thing is that f'(x) is factorizible:

So clearly the roots of f'(x)=0 are x=-1 or x=4/3.
leaving -1. So f(x) has minimum at x=4/3!!!
And here's where the stinky comes:
f(4/3)=823543/6912=119.147...
So for ALL x>0 f(x)>=119.147...!!!
But then
f(a)f(b)f(c)>=119.147...^3=558545864083284007 / 330225942528=1691405.16...
But the main question was to prove that
f(a)f(b)f(c)>=823543 {/*this is 7^7*/}
And ricky an I got that:
f(a)f(b)f(c)>=1691405.16
and the minimum is at {a,b,c}={4/3,4/3,4/3}.
Am I wrong?

krassi_holmz · 2006-03-07 19:17:19

Ganesh, please reply.

Jai Ganesh · 2006-03-07 19:25:37

Sure, krassi_holmz. The problem is being approached in a much different way than I expected. The solution I had to the problem didn't involve differentiation and maximum/minimum. I shall wait for other responses, particularly from mathsyperson/irspow/John. Please wait for the solution for a day more.

krassi_holmz · 2006-03-07 19:43:40

But I'm confused about this: f(a)f(b)f(c)>=7^7
Are there a,b,c for which f(a)f(b)f(c)==7^7?
If there are, so I have a mistake.

Last edited by krassi_holmz (2006-03-08 06:02:48)

Jai Ganesh · 2006-03-08 03:34:22

Solution to ***3

Consider the series 3² /1! + 5² /3! + 7² /5! +..........
the nth term = (2n+1)² /(2n-1)! = (4n² +4n+1)/(2n-1)!
= [(2n-1)(2n-2)+10n-1]/(2n-1)! = 1/(2n-3)! + (10n-1)/(2n-1)!
= 1/(2n-3)! + 5. 1/(2n-2)! + 4. 1/(2n-1)!
Sum to n terms would be

or
5e (on simplification).
Since we started the series as 3² /1! + 5² /3! + 7² /5! +..........,
1 should be added to the sum.

Hence the sum to infinity is 1+5e.

ashwil · 2006-03-08 03:55:35

Many thanks. I had forgotten the technique of considering the nth term. All now clear.

Jai Ganesh · 2006-03-08 16:10:32

***6

Prove that

krassi_holmz · 2006-03-08 19:14:17

***6 beautiful!

so:

Let reduce the left side:

Last edited by krassi_holmz (2006-03-08 19:23:57)

Jai Ganesh · 2006-03-09 03:20:12

Jai Ganesh · 2006-03-09 19:22:51

***7

Let

and
if z be a complex number such that

then prove that |z - 7 -9i| = 3√2.

Jai Ganesh · 2006-03-10 17:43:21

***8

Given that p, q are the roots of the equation Ax²-4x+1=0 and r,s are the roots of the equation Bx²-6x-1=0, find the values of A and B if p, r, q, and s are in Harmonic Progression.

Jai Ganesh · 2006-03-11 18:11:47

***9

Show that if a,b>0, then

Jai Ganesh · 2006-03-12 02:23:36

***10

Show that the curves

and cut orthogonally if .

Jai Ganesh · 2006-03-15 04:24:56

***11

Find the value of a such that the vectors

are coplanar.

Math Is Fun Forum

#26 2006-03-07 01:32:57

Re: *** Problems

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