\(A\ge\frac{1}{2}\left(x-2+3-x\right)^2=\frac{1}{2}\)
\(A_{min}=\frac{1}{2}\) khi \(x-2=3-x\Leftrightarrow x=\frac{5}{2}\)
Hoặc: \(A=x^2-4x+4+x^2-6x+9=2x^2-10x+13=2\left(x-\frac{5}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)
b/
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{b+c+a+c+a+b}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)