Đặt \(\left\{\begin{matrix}a=\frac{1}{3x} & & \\ b=\frac{4}{5y} & & \\c=\frac{3}{2z} \end{matrix}\right.\)\((x,y,z>0)\)
Khi đó \(21a+2bc+8ca\leq12 \Leftrightarrow 3x+5y+7x \leq 15xyz\)
Áp dụng BĐT AM-GM ta có:
\(3x+5y+7z\geq 15\sqrt[15]{x^3y^5z^7}\)
\(\Rightarrow 15xyz\geq 15\sqrt[15]{x^3y^5z^7}=>x^6y^5z^4\geq 1\)
Ta có: \(P = 3x + 2.\dfrac{5}{4}y + 3.\dfrac{2}{3}z \)
\(= \dfrac{1}{2}(6x + 5y + 4z) \ge \dfrac{1}{2}.15\sqrt[{15}]{{{x^6}{y^5}{z^4}}} \ge \dfrac{{15}}{2}\)
Đẳng thức xảy ra khi \(x=y=z=1\) hay \(\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{4}{5}\\c=\dfrac{3}{2}\end{matrix}\right.\)