Cho y ở đề bài làm gì trong khi biểu thức ở vế trái bên dưới ko có y?

à là \(\frac{8x}{y}\)đó

\(VT=3\left(9x^2-12x+4\right)+\frac{8x}{1-x}=27x^2-36x+12+\frac{8x}{1-x}\)

\(=27x^2-36x+4+\frac{8}{1-x}=27x^2-18x-6+8\left(1-x\right)+\frac{8}{1-x}\)

\(=27x^2-18x+3+8\left(1-x\right)+\frac{8}{1-x}-9\)

\(=3\left(3x-1\right)^2+8\left(1-x\right)+\frac{8}{1-x}-9\)

\(\Rightarrow VT\ge2\sqrt{8^2}-9=7\)

Dấu "=" xảy ra khi \(x=\frac{1}{3}\)

\(VT=27x^2-36x+12+\frac{8x}{y}\)

\(=\frac{8x}{1-x}+18x\left(1-x\right)+45x^2-54x+12\)

\(\ge45x^2-54x+12+24x\)

\(=45x^2-30x+12=5\left(9x^2-6x+\frac{12}{5}\right)\)

\(=5\left[\left(3x-1\right)^2+\frac{7}{5}\right]\ge7\)

Dấu = khi \(x=\frac{1}{3};y=\frac{2}{3}\)

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

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

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 \(\left\{{}\begin{matrix}x=\frac{1}{3}\\y=\frac{2}{3}\end{matrix}\right.\)

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

3: \(P=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(z+y\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{y+x}\right)+\dfrac{1}{4}\left(\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)=\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = x = \(\dfrac{1}{3}\).