Question

Cho \(a,b,c>\frac{9}{4}\). Tìm GTNN của \(Q=\frac{a}{2\sqrt{b}-3}+\frac{b}{2\sqrt{c}-3}+\frac{c}{2\sqrt{a}-3}\)

Answer 1

Đặt \(\left\{{}\begin{matrix}2\sqrt{a}-3=x>0\\2\sqrt{b}-3=y>0\\2\sqrt{c}-3=z>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{\left(x+3\right)^2}{4}\\b=\frac{\left(y+3\right)^2}{4}\\c=\frac{\left(z+3\right)^2}{4}\end{matrix}\right.\)

\(A=\frac{\left(x+3\right)^2}{4y}+\frac{\left(y+3\right)^2}{4z}+\frac{\left(z+3\right)^2}{4x}\)

\(Q\ge\frac{\left(x+y+z+9\right)^2}{4\left(x+y+z\right)}=\frac{\left(x+y+z\right)^2+18\left(x+y+z\right)+81}{4\left(x+y+z\right)}\)

\(4Q\ge x+y+z+\frac{81}{x+y+z}+18\ge2\sqrt{\frac{81\left(x+y+z\right)}{x+y+z}}+18=36\)

\(\Rightarrow Q\ge9\Rightarrow Q_{min}=9\) khi \(x=y=z=3\) hay \(a=b=c=9\)