Áp dụng bất đẳng thức Bunhiacopxki ta có :
\(\left[\left(\sqrt{\frac{2}{1-x}}\right)^2+\left(\sqrt{\frac{1}{x}}\right)^2\right].\left[\left(\sqrt{1-x}\right)^2+\left(\sqrt{x}\right)^2\right]\)
\(\ge\left(\sqrt{\frac{2}{1-x}}.\sqrt{1-x}+\sqrt{\frac{1}{x}}.\sqrt{x}\right)^2\)
\(\Rightarrow\left(\frac{2}{1-x}+\frac{1}{x}\right)\left(1-x+x\right)\ge\left(\sqrt{2}+\sqrt{1}\right)^2\)
\(\Rightarrow A\ge3+2\sqrt{2}\)
Dấu " = " xay ra \(\Leftrightarrow\frac{\frac{2}{1-x}}{1-x}=\frac{\frac{1}{x}}{x}\Leftrightarrow\frac{2}{\left(1-x\right)^2}=\frac{1}{x^2}\Leftrightarrow2x^2=\left(1-x\right)^2\)
\(\Leftrightarrow x\sqrt{2}=1-x\left(0< x< 1\right)\)
\(\Leftrightarrow x\left(\sqrt{2}+1\right)=1\Leftrightarrow x=\sqrt{2}-1\)
Chúc bạn học tốt !!
\(A=\frac{1}{x}+\frac{2}{1-x}\ge\frac{\left(1+\sqrt{2}\right)^2}{x+1-x}=3+2\sqrt{2}\)
Dấu "=" xảy ra khi \(x\sqrt{2}=1-x\Leftrightarrow x=\sqrt{2}-1\)
