Đặt \(Z=\frac{2x}{1-x}+\frac{1-x}{x}\)
Áp dụng bđt Cô si với 2 số dương là \(\frac{2x}{1-x}\) và \(\frac{1-x}{x}\) ta có:
\(Z=\frac{2x}{1-x}+\frac{1-x}{x}\ge2.\sqrt{\frac{2x}{1-x}.\frac{1-x}{x}}=2.\sqrt{2}\)
Dấu "=" xảy ra khi \(\frac{2x}{1-x}=\frac{1-x}{x}\)
<=> 2x2 = (1 - x)2 <=> \(\sqrt{2x^2}=\sqrt{\left(1-x\right)^2}\Leftrightarrow\left|x.\sqrt{2}\right|=\left|1-x\right|\)
Mà theo đề bài 0 < x < 1 nên \(\begin{cases}x.\sqrt{2}>0\\1-x>0\end{cases}\)\(\Rightarrow\begin{cases}\left|x.\sqrt{2}\right|=x.\sqrt{2}\\\left|1-x\right|=1-x\end{cases}\)
Do đó, \(x.\sqrt{2}=1-x\Leftrightarrow x.\sqrt{2}+x=1\Leftrightarrow x.\left(\sqrt{2}+1\right)=1\)
\(\Leftrightarrow x=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\)
Xét hiệu: \(y-Z=\left(\frac{2}{1-x}+\frac{1}{x}\right)-\left(\frac{2x}{1-x}+\frac{1-x}{x}\right)=\frac{2-2x}{1-x}+\frac{1-1+x}{x}=2+1=3\)
\(\Leftrightarrow y=Z+3=2.\sqrt{2}+3\)
Vậy Min y = \(2.\sqrt{2}+3\) khi \(x=\sqrt{2}-1\)
