Áp dụng BĐT Cauchy - Schwarz dạng Engel ta có :
\(\dfrac{a^2}{a-1}+\dfrac{b^2}{b-1}\ge\dfrac{\left(a+b\right)^2}{a+b-2}=\dfrac{\left(a+b\right)^2-4+4}{a+b-2}\)
\(=\dfrac{\left(a+b+2\right)\left(a+b-2\right)+4}{a+b-2}=a+b+2+\dfrac{4}{a+b-2}\)
\(=a+b-2+\dfrac{4}{a+b-2}+4\ge2\sqrt{\left(a+b-2\right).\dfrac{4}{a+b-2}}-4=0\)