Có :
\(\left(a^2+4b^2+9c^2\right).\left(1+\frac{1}{4}+\frac{1}{9}\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\frac{49}{36}\ge\left(a+b+c\right)^2\)
\(\Rightarrow A\le\frac{7}{6}\)
c2 : \(\frac{36a^2}{36}+\frac{36b^2}{9}+\frac{36c^2}{4}\ge\frac{\left(6a+6b+6c\right)^2}{49}=\frac{6^2\left(a+b+c\right)^2}{7^2}\)
\(< =>\frac{6^2\left(a+b+c\right)^2}{7^2}\le1< =>a+b+c\le\frac{7}{6}\)