\(A=\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\)
Ta co:\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge2\sqrt{\frac{1}{x}\cdot4x}-3x=4-3x\left(AM-GM\right)\)
Tuong tu:\(y+\frac{1}{y}=4-3y\)
Ta co:\(A\ge\left(4-3x\right)^2+\left(4-3y\right)^2\)
\(=16-24x+9x^2+16-24y+9y^2\)
\(=32-24\left(x+y\right)+9\left(x^2+y^2\right)\)
Ap dung bat dang thuc phu:\(\frac{\left(x+y\right)^2}{4}\le\frac{x^2+y^2}{2}\Rightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)
Khi do,ta co:
\(A\ge32-24\cdot1+9\cdot\frac{1}{2}=\frac{25}{2}\)
Dau bang xay ra khi va chi khi:\(x=y=\frac{1}{2}\)
P/S:E ko chac dau ah,e ms lm quen vs no thoi
\(VT\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\ge\frac{\left(4\left(x+y\right)+\frac{4}{x+y}-3\left(x+y\right)\right)^2}{2}\)
\(\ge\frac{\left(2.4-3.1\right)^2}{2}=\frac{25}{2}\)
đẳng thức xảy ra khi x = y = 1/2