ta có \(x+y=1\Rightarrow x^2+y^2=1-2xy\) theo BĐT Cô si: \(xy\le\dfrac{\left(x+y\right)^2}{4}\Rightarrow2xy\le\dfrac{\left(x+y\right)^2}{2}\Rightarrow1-2xy\ge1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x^2+y^2\ge\dfrac{1}{2}\)
Áp dụng tiếp BĐT Cô Si :\(\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{2}{xy}\ge\dfrac{2}{\dfrac{\left(x+y\right)^2}{4}}=\dfrac{2}{\dfrac{1}{4}}=8\)