a,b,c>0 prove the equation have 1 root :
d^3 - d(ab+bc+ca) - 2abc =0 .
please help me , thanks
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).
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.Originally Posted by math
You didn't have parantheses, so I'll have to assume what you meant.
and such that: .
Prove the inequality
Is that what you meant?