Áp dụng BĐT Bunyakovsky ta có:
\(\left(x+2y\right)^2=\left(x+\sqrt{2}.\sqrt{2}y\right)^2\le\left(1^2+\sqrt{2}^2\right)\left[x^2+\left(\sqrt{2}y\right)^2\right]\)
\(\Leftrightarrow\)\(\left(x+2y\right)^2\le3\left(x^2+2y^2\right)\)
\(\Leftrightarrow\)\(1\le3\left(x^2+2y^2\right)\) (do x + 2y = 1 )
\(\Leftrightarrow\)\(x^2+2y^2\ge\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+2y=1\\\frac{1}{x}=\frac{\sqrt{2}}{\sqrt{2}y}\end{cases}}\)\(\Leftrightarrow\)\(x=y=\frac{1}{3}\)
Vậy \(Min\)\(A=\frac{1}{3}\) \(\Leftrightarrow\)\(x=y=\frac{1}{3}\)
P/s: tham khảo thôi nhé, mk ko chắc đúng (yếu phần cực trị)
\(x^2+2y^2=\left(x+2y\right)^2\) mà \(x+2y=1=>\left(x+2y\right)^2=1^2=1\)
vậy A=1
