จาก $a^2b^2 + b^2c^2 +c^2a^2 =3$, $ab+bc+ca \le 3$, $abc \le 1$

$\displaystyle \sum_{cyc} \sqrt{\dfrac{a+b^2c}{2}}=\sum_{cyc} \sqrt{\dfrac{1}{c}}\sqrt{\dfrac{ac+b^2c^2}{2}} \le \sqrt{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})(\dfrac{ab+bc+ca+3}{2})} = \sqrt{(\dfrac{ab+bc+ca}{abc})(\dfrac{ab+bc+ca+3}{2})} \le \sqrt{\dfrac{9}{abc}} \le \dfrac{3}{abc}$
by Cauchy ครับ