Prove that the function x²+7y²+20z²+8yz-2zx+4xy is never negative for real values to x,y,z?

You can always find a sum of squares for a +def quadratic form like this ...

First write as quadratic in x : x²+2x(2y−z)+(7y²+20z²+8yz)

Complete square on first terms : (x+2y−z)² + (7y²+20z²+8yz) – (2y−z)² = (x+2y−z)² + (3y²+12yz+19z²)

Complete square on last terms : (x+2y−z)² +3(y+2z)²+7z² which is never negative.

Let f = x^2 + 7y^2 + 20z^2 + 8yz - 2zx + 4xy .

Think some terms .

[(2/√5)x + (√5)y]^2 = (4/5)x^2 + 4xy + 5y^2

[(1/√5)x - (√5)z]^2 = (1/5)x^2 - 2xz + 5z^2

[(√2)y + (2√2)z]^2 = 2y^2 + 8yz + 8z^2

So

f = [(2/√5)x + (√5)y]^2 + [(1/√5)x - (√5)z]^2 + [(√2)y + (2√2)z]^2 + 7z^2

4 terms in the right side are all non-negative , so f is non-negative .