Question

Prove that the expressions x2 + (a+b+c)x +a2 +b2 +c2 +2bc –ca –ab can never negative if x,a,b,c are real

Answer 1

I'll admit I used Wolfram Alpha in two major simplification steps.

First, note that the expression is a quadratic polynomial in the variable \(x\). So it can be represented by a upward opening parabola. To ensure that it will never be negative, we need to look at the discriminant, and make sure that the discriminant is always less than or equal to 0.

Answer 2

Since if the discriminant is at most 0, then the parabola will either have no roots (and always be positive) or a single root (and will always be non-negative). Throwing the equation into Wolfram I got the discriminant to be \(-3 a^2+6 a b-3 b^2+6 a c-6 b c-3 c^2\), and simplifying this (again using WA), we obtain that this is equal to \(-3 (a-b-c)^2\). Can you tell me why that expression will always be less than or equal to \(0\)?

Answer 3

@OOOPS that method would also work.

Answer 4

Never negative means discriminant >=0 na?

Answer 5

I think I read your statement a bit wrong. But yes. If your quadratic polynomial is never negative, then it must have a discriminant that is \(\ge0\).

Answer 6

so -3(a-b-c)2 is >= 0? hw?

Answer 7

Today is not a good day for math for me >.<

I got that completely reversed. A quadratic polynomial that is never negative must have discriminant that is always \(\le0\).

Answer 8

ok tthanx buddy,,i'll check it