1.
Ta có:
\(xy\left(x^2+y^2\right)=\dfrac{1}{2}\cdot2xy\left(x^2+y^2\right)\le\dfrac{1}{2}\cdot\dfrac{\left(x^2+2xy+y^2\right)^2}{4}=\dfrac{1}{2}\cdot\dfrac{\left(x+y\right)^4}{4}=\dfrac{1}{2}\cdot\dfrac{2^4}{4}=2\)
\(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{2^2}{4}=1\)
\(\Rightarrow VT\le2\cdot1=2\)
Dấu "=" xảy ra khi x = y = 1