Giải thích các bước giải:
Ta có:
$\dfrac{a+1}{1+b^2}=\dfrac{(a+1)(1+b^2)-b^2(a+1)}{b^2+1}=a+1-\dfrac{b^2(a+1)}{b^2+1}\ge a+1-\dfrac{b^2(a+1)}{2b}=a+1-\dfrac12b(a+1)$
$\to \dfrac{a+1}{1+b^2}\ge a+1-\dfrac12b(a+1)$
Tương tự:
$\dfrac{b+1}{1+c^2}\ge b+1-\dfrac12c(b+1)$
$\dfrac{c+1}{1+a^2}\ge c+1-\dfrac12a(c+1)$
$\to \dfrac{a+1}{1+b^2}+\dfrac{b+1}{1+c^2}+\dfrac{c+1}{1+a^2}\ge (a+b+c+3)-\dfrac12(b(a+1)+c(b+1)+a(c+1))$
$\to \dfrac{a+1}{1+b^2}+\dfrac{b+1}{1+c^2}+\dfrac{c+1}{1+a^2}\ge (a+b+c+3)-\dfrac12(ab+bc+ca+a+b+c)$
$\to \dfrac{a+1}{1+b^2}+\dfrac{b+1}{1+c^2}+\dfrac{c+1}{1+a^2}\ge (a+b+c+3)-\dfrac12(\dfrac13(a+b+c)^2+a+b+c)$
$\to \dfrac{a+1}{1+b^2}+\dfrac{b+1}{1+c^2}+\dfrac{c+1}{1+a^2}\ge (3+3)-\dfrac12(\dfrac13\cdot 3^2+3)$
$\to \dfrac{a+1}{1+b^2}+\dfrac{b+1}{1+c^2}+\dfrac{c+1}{1+a^2}\ge 3$
Dấu = xảy ra khi $a=b=c=1$
