1. ## help me

a,b,c>0 prove the equation have 1 root :
d^3 - d(ab+bc+ca) - 2abc =0 .

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).

3. thansk to rgeq you are very wonderful , and second question
from the equation and x,y,z >0 . ax+by+cz=xyz . Prove the inequality

x+y+z >= 2/d . (\sqrt[(a+d)(b+d)(c+d)])

4. Originally Posted by rgep
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).

the root of equation ???? i don't understand why we get d = sqrt((ab+bc+ca)/3) and d = -sqrt((ab+bc+ca)/2)

5. Typo: it's + or - sqrt((ab+bc+ca)/3)

6. Originally Posted by math
thansk to rgeq you are very wonderful , and second question
from the equation and x,y,z >0 . ax+by+cz=xyz . Prove the inequality

x+y+z >= 2/d . (\sqrt[(a+d)(b+d)(c+d)])

7. x,y,z >0 . ax+by+cz=xyz . with 2/d = 1/(a+d)+1/(b+d)+1/(c+d)Prove the inequality
x+y+z >= 2/d . (\sqrt[(a+d)(b+d)(c+d)])

8. next week i will give the proof to my teacher !!! please

9. Maybe you should make a new thread for your second question because people usually try to answer posts with 0 replies.

10. all member can't help me !!! hix hix hix

11. Seriously,

Start a new thread and more people will see it. Or maybe private message rgep for some more help. Few will be reading this because it is an old thread with many replies.

12. okie i know .
put 2/d = 1/a+d + 1/b+d + 1/c+d
and x,y,z>0 such that : ax+by+cz=xyz . Prove the inequality
(x+y+z)>=\sqrt{(a+d).(b+d).(c+d)} . 2/d

13. Originally Posted by math
okie i know .
put 2/d = 1/a+d + 1/b+d + 1/c+d
and x,y,z>0 such that : ax+by+cz=xyz . Prove the inequality
(x+y+z)>=\sqrt{(a+d).(b+d).(c+d)} . 2/d
Here, I'll do you a favor and put some of this into Latex so people can read it and want to help you more.

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

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

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

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

Is that what you meant?

14. yes , that's right . do you have proof ????

15. it is difficult. but i will try to sovel it. i hope i can sovel this puzzl