Question

[Thịnh]

1. Tìm GTNN Cho x,y là các số thực thỏa mãn x+y=của  P=2x³+x³(2y-1)+y3(2x-1)+2y⁴

Answer 1

[Tran Le Khanh Linh]

Sửa: \(P=2x^4+x^3\left(2y-1\right)+y^3\left(2x-1\right)+2y^4\); x+y=1

Ta có \(P=2x^4+x^3\left(2y-1\right)+y^3\left(2x-1\right)+2y^4=2x^4+2x^3y-x^3+2xy^3-y^3+2y^4\)

\(=x^3\left(2x+2y\right)+y^3\left(2x+2y\right)-\left(x^3+y^3\right)=\left(2x+2y\right)\left(x^3+y^3\right)-\left(x^3+y^3\right)\)

\(=\left(2x+2y-1\right)\left(x^3+y^3\right)=x^3+y^3\)

Do \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=x^2-xy+y^2=\frac{1}{2}\left(x^2+y^2\right)\left(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\right)^2\)

\(\Rightarrow P\ge\frac{1}{2}\left(x^2+y^2\right)\)

Mà \(x+y=1\Rightarrow x^2+y^2+2xy=1\Rightarrow2\left(x^2+y^2\right)-\left(x-y\right)^2=1\)

\(\Rightarrow2\left(x^2+y^2\right)\ge1\Rightarrow\left(x^2+y^2\right)\ge\frac{1}{2}\Rightarrow P\ge\frac{1}{4}\)

Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)