Question

Cho x, y >0 thỏa mãn x+y=1. CMR \(3\left(3x-2\right)^{^2}+\frac{8x}{y}\ge7\)

Answer 1

\(VT=27x^2-36x+12+\frac{15x-7}{1-x}+7\)

\(VT=\frac{-27x^3+63x^2-33x+5}{1-x}+7=\frac{\left(3x-1\right)^2\left(5-3x\right)}{1-x}+7\)

Do \(x< 1\Rightarrow\left\{{}\begin{matrix}5-3x>0\\1-x>0\end{matrix}\right.\) \(\Rightarrow\frac{\left(3x-1\right)^2\left(5-3x\right)}{1-x}\ge0\)

\(\Rightarrow VT\ge7\) (đpcm)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{1}{3}\\y=\frac{2}{3}\end{matrix}\right.\)