Với a , b > 0 . Ta có : \(\left(\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\right)^2\le\left(\sqrt{a}^2+\sqrt{b}^2\right)\left(\sqrt{3a+b}^2+\sqrt{3b+a}^2\right)= \left(a+b\right).4\left(a+b\right)\)
\(\Rightarrow\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}\le2\left(a+b\right)\) ( vì a , b > 0 )
\(\Rightarrow A\ge\frac{1}{2}\left(đpcm\right)\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{3a+b}{a}=\frac{3b+a}{b}\Leftrightarrow a=b\)