Question

\(Q=\dfrac{x+2}{y^2}+\dfrac{y+2}{z^2}+\dfrac{z+2}{x^2}\)

Cho x,y,z>0 thoả mãn x+y+z=xyz.Tìm GTNN của Q

 

Answer 1

\(x+y+z=xyz\Leftrightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)

\(Q=\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+2\left(a^2+b^2+c^2\right)\)

\(Q\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}+2\left(ab+bc+ca\right)=a+b+c+2\)

\(Q\ge\sqrt{3\left(ab+bc+ca\right)}+2=2+\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)