Biểu thức B chỉ có max, ko có min:
Từ giả thiết suy ra \(y^2< 15;z^2< 20\)
\(25x^2+10xyz+20y^2+15z^2=300\)
\(\Leftrightarrow\left(5x+yz\right)^2=y^2z^2-20y^2-15z^2+300\)
\(\Leftrightarrow\left(5x+yz\right)^2=\left(15-y^2\right)\left(20-z^2\right)\le\frac{1}{4}\left(35-y^2-z^2\right)^2\)
\(\Leftrightarrow5x+yz\le\frac{1}{2}\left(35-y^2-z^2\right)\)
\(\Leftrightarrow10x\le35-\left(y+z\right)^2\Rightarrow x\le\frac{35-\left(y+z\right)^2}{10}\)
\(\Rightarrow B\le\frac{35-\left(y+z\right)^2}{10}+y+z=\frac{35-\left(y+z\right)^2+10\left(y+z\right)}{10}=\frac{60-\left(y+z-5\right)^2}{10}\le6\)
\(\Rightarrow B_{max}=6\) khi \(\left(x;y;z\right)=\left(1;2;3\right)\)
