Áp dụng liên tiếp bất đẳng thức Cauchy-Schwarz ta có:
\(\dfrac{a^2+3}{b+c}+\dfrac{b^2+3}{c+a}+\dfrac{c^2+3}{a+b}\)
\(=\dfrac{a^2}{b+c}+\dfrac{3}{b+c}+\dfrac{b^2}{c+a}+\dfrac{3}{c+a}+\dfrac{c^2}{a+b}+\dfrac{3}{a+b}\)
\(=\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)+3\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\)
\(\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}+3.\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}\)
\(=\dfrac{a+b+c}{2}+\dfrac{27}{2\left(a+b+c\right)}=\dfrac{3}{2}+\dfrac{27}{6}=6\)