Question

[ILoveMath]

Cho 0<x,y,z<\(\dfrac{\sqrt{3}}{2}\) thỏa mãn xy+yz+zx=\(\dfrac{3}{4}\)

Tìm Min \(Q=\dfrac{4x^2}{x\left(3-4x^2\right)}+\dfrac{4y^2}{y\left(3-4y^2\right)}+\dfrac{4z^2}{z\left(3-4z^2\right)}\)

Answer 1

[Nguyễn Việt Lâm]

Ta chứng minh BĐT sau:

Ta có: \(x\left(3-4x^2\right)=-4x^3+3x-1+1=1-\left(x+1\right)\left(2x-1\right)^2\le1\)

\(\Rightarrow\dfrac{4x^2}{x\left(3-4x^2\right)}\ge\dfrac{4x^2}{1}=4x^2\)

Tương tự và cộng lại:

\(Q\ge4\left(x^2+y^2+z^2\right)\ge4\left(xy+yz+zx\right)=3\)

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