MQ28 --- Inequality

Let t be any real number.
Then (at+x)²+(bt+y)²+(ct+z)²>=0
(a²t²+2axt+x²)+(b²t²+2byt+y²)+(c²t²+2czt+z²)>=0
(a²+b²+c²)t²+2(ax+by+cz)t+(x²+y²+z²)>=0
The quadratic inequality is non-negative for all real numbers t
⇒[2(ax + by + cz)]²-4 (a² + b² + c²)(x² + y² + z²)≤0
⇒(ax + by + cz)² ≤ (a² + b² + c²)(x² + y² + z²)

(a² + b² + c²)(x² + y² + z²)-(ax + by + cz)²
=(ay-bx)²+(az-cx)²+(bz-cy)²
>= 0
for all x,y,z,a,b,c are real.

Since (a²+b²+c²)t²+2(ax+by+cz)t+(x²+y²+z²)>=0, the discriminant for the quadratic equation in t:
(a²+b²+c²)t²+2(ax+by+cz)t+(x²+y²+z²)=0
should be smaller than or equal to zero, hence:
[2(ax + by + cz)]²-4 (a² + b² + c²)(x² + y² + z²)≤0

題目不對,
試試a=b=c=x=y=z=0.1