[Thgx0312555]

ลองทำดูครับ
by weighted power mean
$$\displaystyle \sum_{cyc}\sqrt{x^2+xy+y^2} = \sum_{cyc} \sqrt{\dfrac{3(x^2+xy+\dfrac{1}{4}y^2)}{3}+\dfrac{3}{4}y^2}$$
$$\ge \displaystyle \sum_{cyc} \dfrac{1}{2}(3\sqrt{\dfrac{x^2+xy+\dfrac{1}{4}y^2}{3}}+\sqrt{\dfrac{3}{4}y^2})$$
$$= \displaystyle \sum_{cyc} \dfrac{1}{2}(\sqrt{3}(x+\dfrac{y}{2})+\dfrac{\sqrt{3}}{2}y)$$
$$= \displaystyle \sum_{cyc}\dfrac{\sqrt{3}}{4}(2x+y+y)$$
$$= \displaystyle \sum_{cyc} \dfrac{\sqrt{3}}{6}(2x+y+3x)$$
$$\ge \displaystyle \sum_{cyc} \dfrac{\sqrt{3}}{6}(2\sqrt{3(2x^2+xy)})$$
$$= \displaystyle \sum_{cyc} \sqrt{2x^2+xy}$$