Question

Cho  2 số thực dương  \(x;y\)  và \(x>y\). Chứng minh rằng \(x+2y+\dfrac{216}{\left(x-y\right).\left(3y+2\right)}\ge16\)

Answer 1

\(A=x+2y+\dfrac{216}{\left(x-y\right)\left(3y+2\right)}=x-y+3y+2+\dfrac{216}{\left(x-y\right)\left(3y+2\right)}-2\)\(\)

\(\Rightarrow x-y+3y+2+\dfrac{216}{\left(x-y\right)\left(3y+2\right)}\ge3\sqrt[3]{\left(x-y\right)\left(3y+2\right).\dfrac{216}{\left(x-y\right)\left(3y+2\right)}}\ge3\sqrt[3]{6^3}\ge18\)

\(\Rightarrow x-y+3y+2+\dfrac{216}{\left(x-y\right)\left(3y+2\right)}-2\ge18-2\ge16\)

\(\Rightarrow A\ge16\left(dpcm\right)\) \(dấu"="\) \(xảy\) \(ra\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{22}{3}\\y=\dfrac{4}{3}\end{matrix}\right.\)