Giải thích các bước giải:
2.Ta có : $(a+b+c)^2\ge 3(ab+bc+ca)=3\rightarrow a+b+c\ge\sqrt{3}$
Ta có :
$b^2+1=\dfrac{3}{4}.(b^2+1)(\dfrac{1}{3}+1)\ge \dfrac{3}{4}(\dfrac{b}{\sqrt{3}}+1)^2$
$\rightarrow a\sqrt{b^2+1}\ge a.\sqrt{\dfrac{3}{4}(\dfrac{b}{\sqrt{3}}+1)^2}=\dfrac{\sqrt{3}}{2}.a(\dfrac{b}{\sqrt{3}}+1)$
Tương tự ta chứng minh được
$ b\sqrt{c^2+1}\ge \dfrac{\sqrt{3}}{2}.b(\dfrac{b}{\sqrt{3}}+1)$
$ c\sqrt{a^2+1}\ge \dfrac{\sqrt{3}}{2}.c(\dfrac{a}{\sqrt{3}}+1)$
$\rightarrow a\sqrt{b^2+1}+b\sqrt{c^2+1}+ c\sqrt{a^2+1}\ge \dfrac{\sqrt{3}}{2}.a(\dfrac{b}{\sqrt{3}}+1)+ \dfrac{\sqrt{3}}{2}.b(\dfrac{b}{\sqrt{3}}+1)+\dfrac{\sqrt{3}}{2}.c(\dfrac{a}{\sqrt{3}}+1)$
$\rightarrow a\sqrt{b^2+1}+b\sqrt{c^2+1}+ c\sqrt{a^2+1}\ge \dfrac{1}{2}(ab+bc+ca)+\dfrac{\sqrt{3}}{2}(a+b+c)$
$\rightarrow a\sqrt{b^2+1}+b\sqrt{c^2+1}+ c\sqrt{a^2+1}\ge \dfrac{1}{2}.1+\dfrac{\sqrt{3}}{2}.\sqrt{3}$
$\rightarrow a\sqrt{b^2+1}+b\sqrt{c^2+1}+ c\sqrt{a^2+1}\ge 2$