Với a,b>0 ta có ($\sqrt[]{a(3a+b)}$ +$\sqrt[]{b(3b+a)})^2$ $\leq$ ($\sqrt[]{a}^2$ +$\sqrt[]{b}^2$ )($\sqrt[]{3a+b}^2$+ $\sqrt[]{3b+a}^2$) =(a+b).4(a+b)
=>$\sqrt[]{a(3a+b)}$ +$\sqrt[]{b(3b+a)}$ $\leq$ 2(a+b)(do a,b>0)
=>A $\geq$ $\frac{1}{2}$ Đpcm
Dấu = xảy ra <=> $\frac{3a+b}{a}$= $\frac{3b+a}{b}$ <=>a=b