-Áp dụng BĐT Schwarz cho 3 số thực dương ta có:
\(\dfrac{a^2}{b+3c}+\dfrac{b^2}{c+3a}+\dfrac{c^2}{a+3b}\ge\dfrac{\left(a+b+c\right)^2}{b+3c+c+3a+a+3b}=\dfrac{\left(a+b+c\right)^2}{4\left(a+b+c\right)}=\dfrac{a+b+c}{4}\)-Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

