If a,b,c are three positive real numbers such that a+b+c=1, and

N=min(a^3+[a^2]bc, b^3+[b^2]ac, c^3+[c^2]ab),

Then PT the roots of the equation x^2+x+4N=0, are real.

Hi Speed Racer,

There's a nice concept you can get here. Each term is of the form x³+x(abc), where x= a or b or c.

Say a≥b≥c, then a³+a²bc ≥ b³+b²ac ≥ c³+c²ab.

So min value will be c³+c²ab. And note that c cannot be greater than 1/3 (as a+b+c=1, and c is not greater than a or b).

Also abc is maximum when a=b=c=1/3 for a+b+c=1.

So N can take the maximum value, when a=b=c=1/3.

Consider the discriminant of the QE, we have D = 1-16N.

Lets see what's the max value that N can take. ie when a=b=c=1/3.

And that value is (1/3)^3 + (1/3)^4 = 4/81 which is always less than 1/16.

So 1-16N is always positive, and hence the roots of the QE are always real, and that solves the problem.

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).