Cho $a,b,c$ là các số thực dương. CMR:
$\sum\dfrac{a}{b+c}+\sqrt{\dfrac{ab+bc+ca}{a^2+b^2+c^2}}\geq \dfrac{5}{2}$

$\sum \dfrac{a}{b+c}\geq \dfrac{(a+b+c)^2}{2(ab+bc+ac)}=\dfrac{a^2+b^2+c^2}{2ab+2bc+2ac}+1$
BĐT <=> $VT\geq \dfrac{1}{2}\left ( \dfrac{a^2+b^2+c^2}{ab+bc+ac} \right )+1+\sqrt{\dfrac{ab+bc+ac}{a^2+b^2+c^2}}=\dfrac{1}{2}.\left ( \dfrac{\sum a^2}{\sum ab} \right )+\dfrac{1}{2}\sqrt{\left ( \dfrac{\sum ab}{\sum a^2} \right )}+\dfrac{1}{2}\sqrt{\left ( \dfrac{\sum ab}{\sum a^2} \right )}+1\geq \dfrac{3}{2}+1=\dfrac{5}{2}$