Question

Cho a, b là các số thực dương thỏa mãn \(\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=4\)

Tìm GTNN của \(\frac{a^2}{b}+\frac{b^2}{a}\)

Answer 1

Đặt \(\left(\sqrt{a};\sqrt{b}\right)=\left(x;y\right)\)

\(\left(x+1\right)\left(y+1\right)=4\Leftrightarrow3=xy+x+y\le\frac{1}{4}\left(x+y\right)^2+x+y\)

\(\Rightarrow\left(x+y\right)^2+4xy-12\ge0\)

\(\Leftrightarrow\left(x+y+6\right)\left(x+y-2\right)\ge0\)

\(\Leftrightarrow x+y-2\ge0\Rightarrow x+y\ge2\)

\(P=\frac{x^4}{y^2}+\frac{y^4}{x^2}\ge\frac{\left(x^2+y^2\right)^2}{x^2+y^2}=x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\ge\frac{1}{2}.4=2\)

\(P_{min}=2\) khi \(x=y=1\) hay \(a=b=1\)