Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

***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!

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

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.

Character is who you are when no one is looking.

Offline

**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

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!

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

***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)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Logaritms?

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Can someone do this:

7(ln(a + 1) + ln(b + 1) + ln(c + 1) - ln 7) ≥ 4ln abc

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Well done, Ricky!

This was great idea!

Your function has local minimum between 1 and 2 at x=4/3.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

***5

Something stinks here.

And where's the ricky's post?

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Let

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Here's a plot:

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

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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?

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Ganesh, please reply.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

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.

Character is who you are when no one is looking.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

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)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

somewhere...

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

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.

Character is who you are when no one is looking.

Offline

**ashwil****Member**- Registered: 2006-02-27
- Posts: 121

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

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

***6

Prove that

Character is who you are when no one is looking.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

***6 beautiful!

so:

Let reduce the left side:

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

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

Character is who you are when no one is looking.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

***7

Let

and

if z be a complex number such that

then prove that |z - 7 -9i| = 3√2.

Character is who you are when no one is looking.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

***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.

Character is who you are when no one is looking.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

***9

Show that if a,b>0, then

Character is who you are when no one is looking.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,608

***10

Show that the curves

and cut orthogonally if .Character is who you are when no one is looking.

Offline