\(A=\dfrac{3a}{b+c}+\dfrac{4b}{c+a}+\dfrac{3c}{a+b}\)
Đặt \(\left\{{}\begin{matrix}a+b=x\\b+c=y\\c+a=z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{x+z-y}{2}\\b=\dfrac{x+y-z}{2}\\c=\dfrac{y+z-x}{2}\end{matrix}\right.\)
\(\Rightarrow2A=\dfrac{3\left(x+z-y\right)}{y}+\dfrac{4\left(x+y-z\right)}{z}+\dfrac{5\left(y+z-x\right)}{x}\)
\(=-12+\left(\dfrac{3x}{y}+\dfrac{5y}{x}\right)+\left(\dfrac{4y}{z}+\dfrac{3z}{y}\right)+\left(\dfrac{4x}{z}+\dfrac{5z}{x}\right)\ge-12+2\sqrt{15}+4\sqrt{3}+4\sqrt{5}\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3x}{y}=\dfrac{5y}{x}\\\dfrac{4y}{z}=\dfrac{3z}{y}\\\dfrac{4x}{z}=\dfrac{5z}{x}\end{matrix}\right.\)
Ace Legona , Xuân Tuấn Trịnh , Nguyễn Huy Tú ,Hung nguyen,..... làm giúp đi , còn nhiều lắm luôn ......
