Đặt \(\left\{{}\begin{matrix}b+c-a=2x\\c+a-b=2y\\a+b-c=2z\end{matrix}\right.\)\(\forall x,y,z>0\Rightarrow\left\{{}\begin{matrix}a=y+z\\b=x+z\\c=x+y\end{matrix}\right.\)
Khi đó: \(S=\dfrac{y+z}{2x}+\dfrac{4\left(x+z\right)}{2y}+\dfrac{9\left(x+y\right)}{2z}\)
\(\Rightarrow2S=\dfrac{y+z}{x}+\dfrac{4\left(x+z\right)}{y}+\dfrac{9\left(x+y\right)}{z}\)
\(=\left(\dfrac{y}{x}+\dfrac{4x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{9x}{z}\right)+\left(\dfrac{4z}{y}+\dfrac{9y}{z}\right)\)
Áp dụng BĐT AM-GM ta có:
\(2S\ge2\sqrt{\dfrac{y}{x}\cdot\dfrac{4x}{y}}+2\sqrt{\dfrac{z}{x}\cdot\dfrac{9x}{z}}+2\sqrt{\dfrac{4z}{y}+\dfrac{9y}{z}}\)
\(\Leftrightarrow2S\ge2\sqrt{4}+2\sqrt{9}+2\sqrt{36}\)
\(\Leftrightarrow2S\ge4+6+12=22\Leftrightarrow S\ge11\)
