Ta có:$\frac{a}{1+b^{2}c}=a-\frac{b^{2}c}{1+b^{2}c}\geq a-\frac{b\sqrt{c}}{2}$.
Tương tự rồi cộng vế vs vế ta có:
$VT\geq (a+b+c)-\frac{b\sqrt{c}+c\sqrt{a}+a\sqrt{b}}{2}=3-\frac{b\sqrt{c}+c\sqrt{a}+a\sqrt{b}}{2}$.
BĐT cần cm <-->$b\sqrt{c}+c\sqrt{a}+a\sqrt{b}\leq3$.$VT \Leftrightarrow \sqrt{b}.\sqrt{bc}+\sqrt{c}\sqrt{ca}+\sqrt{a}\sqrt{ab}\leq \sqrt{(a+b+c)(ab+bc+ca)}\leq \sqrt{(a+b+c)(\frac{(a+b+c)^2}{3})}=3$