IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

I think the 1st part can be proved in an easier manner:

Proving that $abc \ne 0$ is dealt in the same manner as above as otherwise it contradicts $\beta \ne 0$

Now we have the following inequalities:

1. a+b+c = $(\alpha + i\beta) + (\alpha - i\beta) = 2\alpha >0$

2. ab+bc+ca = $\alpha^2+\beta^2 > 0$

3.

Ineq 3 obviates the need for ineq 2 and so we deal only with the inequalities

and

Now, Ineq 1 means we cannot have all of a,b and c negative.

We are left with the task of eliminating the cases when only two are +ve or only two are -ve

Now, $a^2+b^2+c^2 <2(ab+bc+ca) \Rightarrow (a-b)^2 < 2c(a+b) - c^2$

$\Rightarrow c[2(a+b) - c] >0$

Now, suppose c>0 and a,b<0

Then $2(a+b) - c<0 \ \text{and} \ c>0$ contradicting

Again if c<0 and a,b>0 we have

$2(a+b)-c >0 \ \text{and} c<0$

and we again obtain a contradiction.

Hence the only possibility is a,b and c are all +ve