Question

4) Cho x,y > 0 ; x + y = 1 . Tìm min M = \(\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}\)

Answer 1

\(M=\dfrac{1}{x^{2}+y^{2}}+\dfrac{1}{xy} \\=(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy})+\dfrac{1}{2xy}\\ \)

\(\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{2.\left(\dfrac{x+y}{2}\right)^2}=\dfrac{4}{1^2}+\dfrac{1}{2.\left(\dfrac{1}{2}\right)^2}=6\)

Dấu "=" xảy ra<=>x=y=0,5.

Answer 2

\(M=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{\left(1+1\right)^2}{x^2+y^2+2xy}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=6\)

\(\Rightarrow M_{min}=6\) khi \(x=y=\dfrac{1}{2}\)