Question

Cho \(\hept{\begin{cases}x+y\le1\\x,y>0\end{cases}}\)

Tìm MIN A=\(\left(\frac{1}{x}+\frac{1}{y}\right)\sqrt{1+x^2y^2}\)

Answer 1

Vì a,b>0

A\(\ge2\sqrt{\frac{1}{x}\cdot\frac{1}{y}}\cdot\sqrt{1+x^2y^2}\)

A\(\ge2\sqrt{\frac{1+x^2y^2}{xy}}\)

A\(\ge2\sqrt{\frac{1}{xy}+xy}\)

Đặt xy=a, a>0

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

ĐK 0<a<\(\frac{1}{4}\)

\(\Leftrightarrow A\ge2\sqrt{\frac{1}{a}+a}\)

A\(\ge2\sqrt{16a+\frac{1}{a}-15a}\)

a>0, áp dụng bđt cô si

\(A\ge2\sqrt{2\sqrt{16a\cdot\frac{1}{a}}-\frac{15}{4}}\)

A\(\ge\sqrt{17}\)

Dấu = x ra a=b=0.5