Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+2^2\right)\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge\left(x+4y\right)^2\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge1^2=1\)
\(\Rightarrow5\left(x^2+4y^2\right)\ge\dfrac{1}{5}\)
Đẳng thức xảy ra khi \(x=y=\dfrac{1}{5}\)