\(\dfrac{-1}{4}x^2+x-3=-\left(\dfrac{1}{4}x^2-x+3\right)\)
\(=-\left(\dfrac{1}{4}x^2-\dfrac{1}{2}.x.2+1+2\right)\)
\(=-\left[\left(\dfrac{1}{2}x-1\right)^2+2\right]=-\left(\dfrac{1}{2}x-1\right)^2-2\le-2\)
\(\Rightarrow\)Biểu thức trên luôn âm ( đpcm )