[tex]\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{1}{2}(a+b)^2+\frac{3}{2}(a^2+b^2)}\geq \sqrt{\frac{3}{4}(a+b)^2+\frac{1}{2}(a+b)^2}=\frac{\sqrt{5}}{2}(a+b)[/tex]
tương tự với [tex]\sqrt{2b^2+bc+2c^2};\sqrt{2c^2+ca+2a^2}[/tex] => [tex]P\geq \sqrt{5}(a+b+c)\geq \frac{\sqrt{5}}{3}(\sqrt{a}+\sqrt{b}+\sqrt{c})=\frac{\sqrt{5}}{3}[/tex] dấu = xảy ra khi a=b=c=1/9