Đặt \(x=b+c-a,y=a+c-b,z=a+b-c\) . Khi đó x,y,z >0 và \(a=\frac{y+z}{2},b=\frac{x+z}{2},c=\frac{x+y}{2}\)
Vậy \(P=\frac{2y+2z}{x}+\frac{9x+9z}{2y}+\frac{8x+8y}{z}=\left(\frac{2y}{x}+\frac{9x}{2y}\right)+\left(\frac{2z}{x}+\frac{8x}{z}\right)+\left(\frac{9z}{2y}+\frac{8y}{z}\right)\)
\(\ge2\sqrt{9}+2\sqrt{16}+2\sqrt{36}\). Dấu '=' xảy ra khi:
\(\hept{\begin{cases}\frac{2y}{x}=\frac{9x}{2y}\\\frac{2z}{x}=\frac{8x}{z}\\\frac{9z}{2y}=\frac{8y}{z}\end{cases}\Leftrightarrow\hept{\begin{cases}4y^2=9x^2\\2z^2=8x^2\\9z^2=8y^2\end{cases}}}\Leftrightarrow\hept{\begin{cases}x,y,z>0\\2x=z\\2y=3x;3z=4y\end{cases}}\)
