20.ĐK: $a,b,c> 0$; $a+b+c=3$
$\frac{a+1}{b^{2}+1}+\frac{b+1}{c^{2}+1}+\frac{c+1}{a^{2}+1}\geq 3$
$\frac{a+1}{b^2+1}=a+1-\frac{b^2(a+1)}{b^2+1}\geq a+1-\frac{b^2(a+1)}{2b}=a+1-\frac{ab+b}{2}$
Tương tự $\Rightarrow VT\geq 3+\frac{\sum a}{2}-\frac{\sum ab}{2}\geq 4.5-\frac{(\sum a)^2}{6}=VP(đpcm)$
Dấu ''='' xr khi a=b=c=1
- Element hero Neos và DangHongPhuc thích