a,b,c>0 prove the equation have 1 root :

d^3 - d(ab+bc+ca) - 2abc =0 .

please help me , thanks

Results 1 to 15 of 15

- October 15th 2005, 03:30 PM #1

- Joined
- Oct 2005
- Posts
- 17

- October 15th 2005, 10:47 PM #2
If a cubic f has three real roots then it has a (real) maximum and a minimum between the roots. So the derivative f' has to have two real roots, and the value of f at the maximum must be positive and the value at the minimum must be negative. In this case f' is 3d^2 - (ab+bc+ca). At d = sqrt((ab+bc+ca)/3) we have f = d(d^2 - (ab+bc+ca)) - 2abc = -(2/3)(ab+bc+ba)^(3/2) - 2abc and this is indeed negative. At d = -sqrt((ab+bc+ca)/2) we have f = d(d^2 - (ab+bc+ca)) - 2abc = (2/3)(ab+bc+ba)^(3/2) - 2abc. Now (ab+bc+bc)/3 >= (ab.bc.ca)^(1/3) by the AM-GM inequality and this shows that the value at the maximum is negative unless there is equality in the AM-GM which happens only when a,b,c are all equal. We conclude that if a=b=c there are three real roots, at -a,-a and 2a; if a,b,c are not all equal there is one real root which lies to the right of sqrt((ab+bc+ca)/3).

- October 15th 2005, 10:58 PM #3

- Joined
- Oct 2005
- Posts
- 17

- October 17th 2005, 01:59 AM #4

- Joined
- Oct 2005
- Posts
- 17

- October 17th 2005, 12:09 PM #5

- October 20th 2005, 08:23 PM #6

- Joined
- Oct 2005
- Posts
- 17

- October 20th 2005, 08:27 PM #7

- Joined
- Oct 2005
- Posts
- 17

- October 26th 2005, 02:44 PM #8

- Joined
- Oct 2005
- Posts
- 17

- October 26th 2005, 03:01 PM #9

- October 27th 2005, 05:16 AM #10

- Joined
- Oct 2005
- Posts
- 17

- October 27th 2005, 06:29 AM #11

- October 27th 2005, 02:40 PM #12

- Joined
- Oct 2005
- Posts
- 17

- October 27th 2005, 03:53 PM #13

- Joined
- Oct 2005
- From
- Earth
- Posts
- 1,599

Originally Posted by**math**

You didn't have parantheses, so I'll have to assume what you meant.

Put

and such that: .

Prove the inequality

Is that what you meant?

- October 27th 2005, 08:03 PM #14

- Joined
- Oct 2005
- Posts
- 17

- January 15th 2006, 01:11 AM #15

- Joined
- Jan 2006
- Posts
- 2