Question

Cho x,y là số thực dương.Chứng minh rằng :

\(\left(x+y\right)^2+\dfrac{x+y}{2}\ge2x\sqrt{y}+2y\sqrt{x}\)

Giúp mk nha mn

Answer 1

\(\Leftrightarrow\left(x+y\right)\left(x+y+\dfrac{1}{2}\right)\ge2x\sqrt{y}+2y\sqrt{x}\\ \Leftrightarrow2\sqrt{xy}\left(x+y+\dfrac{1}{2}\right)\ge2\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\)

\(\Leftrightarrow x-\sqrt{x}+\dfrac{1}{4}+y-\sqrt{y}+\dfrac{1}{4}\ge0\\ \Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\left(\sqrt{y}-\dfrac{1}{2}\right)^2\ge0\)

Dấu = xảy ra khi x=y=1/4