Đáp án:
\[A_{\min}=\dfrac{27}5\Leftrightarrow a= b=c=\dfrac 13\]
Giải thích các bước giải:
\(A=\dfrac 1{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}\ge \dfrac{9}{2a-a^2+2b-b^2+2c-c^2}\\\to A\ge \dfrac{9}{2\left(a+b+c\right)-\left(a^2+b^2+c^2\right)}\\\to A\ge \dfrac{9}{2-\left(a^2+b^2+c^2\right)}\ge \dfrac{9}{2-\left(\dfrac{\left(a+b+c\right)^2}{3}\right)}=\dfrac{9}{2-\dfrac 13}=\dfrac{27}5\)
\(\to A_{\min}=\dfrac{27}5\Leftrightarrow a= b=c=\dfrac 13\)