Question

Cho \(x,y\) là các số thực thỏa mãn \(2x^2+y^2+xy\ge1\). Tìm GTNN của biểu thức: \(A=x^2+y^2\)

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Answer 1

\(\Rightarrow\dfrac{A}{2x^2+y^2+xy}\le\dfrac{A}{1}\Leftrightarrow\dfrac{x^2+y^2}{2x^2+y^2+xy}\le A\)

Đặt \(\dfrac{x}{y}=t\). Ta có:

\(P=\dfrac{x^2+y^2}{2x^2+y^2+xy}=\dfrac{\left(\dfrac{x^2}{y^2}\right)+\left(\dfrac{y^2}{y^2}\right)}{\left(\dfrac{2x^2}{y^2}\right)+\left(\dfrac{y^2}{y^2}\right)+\left(\dfrac{xy}{y^2}\right)}=\dfrac{t^2+1}{2t^2+1+t}\)

\(\Rightarrow2t^2P+P+Pt=t^2+1\Leftrightarrow t^2\left(2P-1\right)+Pt+P-1\)

\(\Delta=P^2-4\left(2P-1\right)\left(P-1\right)\ge0\)

\(\Rightarrow\dfrac{6-2\sqrt{2}}{7}\le P\le\dfrac{6+2\sqrt{2}}{7}\)

\(\Rightarrow A\ge P\ge\dfrac{6-2\sqrt{2}}{7}\)