Question

1.Cho x,y,z >0 thoả mãn : x(3-xy-xz) + y +6z =< 5xz(y+z)

GTNN của P = 3x + y + 6z 

Answer 1

2x+80=3y

Answer 2

\(3x+y+6z\le x^2\left(y+z\right)+5xz\left(y+z\right)=\dfrac{1}{2}.2x\left(y+z\right)\left(x+5z\right)\)

\(\Rightarrow3x+y+6z\le\dfrac{1}{54}\left(2x+y+z+x+5z\right)^3=\dfrac{1}{54}\left(3x+y+6z\right)^3\)

\(\Rightarrow\left(3x+y+6z\right)^2\ge54\)

\(\Rightarrow3x+y+6z\ge3\sqrt{6}\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{\sqrt{6}}{2};\dfrac{9\sqrt{6}}{10};\dfrac{\sqrt{6}}{10}\right)\)