Question

[Lizy]

`x,y,z` là các số thực lớn hơn `0`, `4x+7y+6z >= 76`

Tìm min \(P=x+y+z+\dfrac{9}{2x}+\dfrac{2}{y}+\dfrac{9}{z}\)

Answer 1

[Nguyễn Việt Lâm]

\(P=\left(\dfrac{x}{2}+\dfrac{9}{2x}\right)+\left(\dfrac{y}{8}+\dfrac{2}{y}\right)+\left(\dfrac{z}{4}+\dfrac{9}{z}\right)+\dfrac{1}{8}\left(4x+7z+6z\right)\)

\(P\ge2\sqrt{\dfrac{9x}{4x}}+2\sqrt{\dfrac{2y}{8y}}+2\sqrt{\dfrac{9z}{4z}}+\dfrac{1}{8}.76=\dfrac{33}{2}\)

Dấu "=" xảy ra tại \(\left(x;y;z\right)=\left(3;4;6\right)\)