\(\left(\frac{47}{12}\right)^2=\left(a+b+c\right)^2=\left(\frac{1}{\sqrt{3}}.\sqrt{3}a+\frac{1}{2}.2b+\frac{1}{\sqrt{5}}.\sqrt{5}c\right)^2\)
\(\Rightarrow\left(\frac{47}{12}\right)^2\le\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\left(3a^2+4b^2+5c^2\right)\)
\(\Rightarrow A\ge\frac{\left(\frac{47}{12}\right)^2}{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}}=\frac{235}{12}\)
\(A_{min}=\frac{235}{12}\) khi \(\left\{{}\begin{matrix}a+b+c=\frac{47}{12}\\3a=4b=5c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{5}{3}\\b=\frac{5}{4}\\c=1\end{matrix}\right.\)
