Question

Cho x,y > 0 và x + y = 1 . Tìm min của \(S=\dfrac{1}{x^2+y^2}\) + \(\dfrac{5}{xy}\)

Answer 1

Áp dụng BĐT AM-GM ta có:

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

Áp dụng BĐT Cauchy-Schwarz ta có:

\(S=\dfrac{1}{x^2+y^2}+\dfrac{5}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{9}{2xy}\)

\(\ge\dfrac{\left(1+1\right)^2}{x^2+2xy+y^2}+\dfrac{9}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{9}{2\cdot\dfrac{1}{4}}=22\)

Xảy ra khi \(x=y=\dfrac{1}{2}\)