Prove that the expression x^2 - (a+b+c)x+a^2+b^2+c^2+2bc -ca-ab can never be negative if x,a,b,c, are real.

There are two proofs to it

1) let us multiply by 4 to make it denominator free

4x^2 - 4(a+b+c)x+4(a^2+b^2+c^2+2bc -ca-ab)

= (2x - (a+b+c))^2 + 4(a^2+b^2+c^2+2bc -ca-ab) - (a+b+c)^2

= (2x - (a+b+c))^2 + 4(a^2+b^2+c^2+2bc -ca-ab) - (a^2+b^2 + c^2 + 2ab + 2bc + 2ca)

= (2x - (a+b+c))^2 + (3a^2+3b^2+3c^2+6bc -6ca-6ab)

= (2x - (a+b+c))^2 + 3(a^2+b^2+c^2+2bc -2ca-2ab)

= (2x - (a+b+c))^2 + 3(a^2+(b+c)^2 -2a(b+c))
= (2x - (a+b+c))^2 + 3(a-(b+c))^2

as it it sum of squares it cannot be -ve

Method 2

Using the quadratic formula gives:

x = (-(-(a+b+c) ± √((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)))/2

So if ((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)) < 0 it will only have complex roots and so wont cross the x axis. Since the coefficient of x² is positive this means it would only be positive. Therefore one need to prove ((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)) < 0

o r

-3a² -3b² - 3c² + 6ab + 6ac - 6bc < 0

factoring gives
-3(a-b-c)² < 0

which is true unless a-b-c = 0 it will have complex roots and so only be positive.

If a-b-c = 0 that means it has equal roots and so will only touch the x axis without crossing it and so will only be positive or 0.