Question

Answer 1

\(ab+3c=ab+\left(a+b+c\right)c=\left(a+c\right)\left(b+c\right)\)

\(\Rightarrow A=\dfrac{\sqrt{\left(a+c\right)\left(b+c\right)}+\sqrt{2\left(a^2+b^2\right)}}{3+\sqrt{ab}}\ge\dfrac{\sqrt{\left(\sqrt{ab}+c\right)^2}+\sqrt{\left(a+b\right)^2}}{3+\sqrt{ab}}\)

\(\Rightarrow A\ge\dfrac{\sqrt{ab}+c+a+b}{3+\sqrt{ab}}=\dfrac{\sqrt{ab}+3}{3+\sqrt{ab}}=1\)

\(A_{min}=1\) khi \(a=b=c=1\)