Có \(\sqrt{\left(3a+b\right)\left(a+3b\right)}\le\frac{3a+b+a+3b}{2}=2\left(a+b\right)\)
Mà 4ab=\(\left(2\sqrt{ab}\right)^2=\left[\left(\sqrt{a}+\sqrt{b}\right)^2-\left(a+b\right)\right]^2=\left[1-\left(a+b\right)\right]^2\)
Do đó nếu đặt a+b=t. Khi đó a+b \(\ge\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2}=\frac{1}{2}\)
hay \(t\ge\frac{1}{2}\)
Cần chứng minh: \(3\left(a+b\right)^2-\left(a+b\right)+4ab\ge\frac{1}{2}\sqrt{\left(3a+b\right)\left(a+3b\right)}\)
\(\Leftrightarrow3t^2-t+\left(1-t\right)^2\ge\frac{1}{2}\cdot2t\)
\(\Leftrightarrow4t^2-4t+1\ge0\)
\(\Leftrightarrow\left(2t-1\right)^2\ge0\)luôn đúng với mọi t \(\ge\frac{1}{2}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2t-1=0\\3a+b=3b+a\\\sqrt{a}+\sqrt{b}=1\end{cases}\Leftrightarrow\hept{\begin{cases}a+b=\frac{1}{2}\\a=b\\\sqrt{a}+\sqrt{b}=1\end{cases}\Leftrightarrow}a=b=\frac{1}{4}}\)
