-BĐT Caushy:\(4xy\le\left(x+y\right)^2\Leftrightarrow4xy\le4\Leftrightarrow xy\le1\)
\(xy\left(x+y\right)=2xy\le2.1=2\)
-Dấu bằng xảy ra \(\Leftrightarrow x=y=1\)
-Sửa lại:
\(4xy\le\left(x+y\right)^2\Leftrightarrow4xy\le4\Leftrightarrow xy\le1\)
\(2xy\left(x^2+y^2\right)\le2xy.2xy=4x^2y^2\)
-Ta c/m: \(4x^2y^2\le\left(x+y\right)^2\)
\(\Leftrightarrow4x^2y^2\le x^2+y^2+2xy\)
\(\Rightarrow4\le\dfrac{1}{x^2}+\dfrac{2}{xy}+\dfrac{1}{y^2}\)
\(\Leftrightarrow4\le\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Leftrightarrow4\le\left(\dfrac{2}{xy}\right)^2\) (đúng do \(xy\le1\)
\(\Rightarrow2xy\left(x^2+y^2\right)\le\left(x+y\right)^2=4\)
\(\Leftrightarrow xy\left(x^2+y^2\right)\le4\) (đpcm)
-Dấu bằng xảy ra \(\Leftrightarrow x=y=1\)
