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

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

please help me , thanks

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

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- Oct 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).

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

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

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

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

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

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

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

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

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Originally Posted by**math**

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

Put $\displaystyle \frac{2}{d}=\frac{1}{a+d}+\frac{1}{b+d}+\frac{1}{c +d}$

and $\displaystyle x,y,z>0$ such that: $\displaystyle ax+by+cz=xyz$.

Prove the inequality $\displaystyle x+y+z\ge \frac{2\sqrt{(a+d)(b+d)(c+d)}}{d}$

Is that what you meant?

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

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

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