Question

Cho x + y \(\le\)1 .Tìm Min B = \(\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\)

Answer 1

\(B=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\)

\(\ge\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\dfrac{\left(1+0,25xy\right)}{\sqrt{1,0625}}\)

\(=\dfrac{1}{\sqrt{1,0625}}\left(\dfrac{1}{x}+\dfrac{1}{y}+0,25x+0,25y\right)\)

\(=\dfrac{1}{\sqrt{1,0625}}\left(\left(\dfrac{1}{x}+4x\right)+\left(\dfrac{1}{y}+4y\right)-\dfrac{15}{4}\left(x+y\right)\right)\)

\(\ge\dfrac{1}{\sqrt{1,0625}}\left(4+4-\dfrac{15}{4}\right)=\sqrt{17}\)

Answer 2

\(\sqrt[1]{\sqrt[1]{1}}15\)