Question

[Phác Chí Mẫn]

Cho x, y, z dương thỏa \(x+y+z=\frac{3}{2}\). Tìm min: \(P=\frac{\sqrt{x^2+xy+y^2}}{1+4xy}+\frac{\sqrt{z^2+zy+y^2}}{1+4zy}+\frac{\sqrt{x^2+xz+z^2}}{1+4xz}\)

Answer 1

[Nguyễn Việt Lâm]

\(x^2+xy+y^2=\left(x+y\right)^2-xy\ge\left(x+y\right)^2-\frac{1}{4}\left(x+y\right)^2=\frac{3}{4}\left(x+y\right)^2\)

\(\Rightarrow\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\)

Vậy:

\(P\ge\frac{\sqrt{3}}{2}\left[\frac{\left(x+y\right)^2}{1+4xy}+\frac{\left(y+z\right)^2}{1+4yz}+\frac{\left(z+x\right)^2}{1+4zx}\right]\)

\(P\ge\frac{\sqrt{3}}{2}\left[\frac{\left(2x+2y+2z\right)^2}{3+4\left(xy+yz+zx\right)}\right]\ge\frac{\sqrt{3}}{2}.\frac{9}{3+\frac{4}{3}\left(x+y+z\right)^2}=\frac{3\sqrt{3}}{4}\)

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

Answer 2

[Phác Chí Mẫn]

\(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}{2}\left(x+y\right)\) mà sao thế vào là \(\frac{\sqrt{3}}{2}\left(x+y\right)^2\) vậy ạ?