[tex]UCT:2a^2+2b^2+ab\geq \frac{5}{4}(a+b)^2[/tex]
Có:
[tex]P=\sum \sqrt{2a^2+ab+2b^2}\geq \sum \sqrt{\frac{5}{4}(a+b)^2}=\sum \frac{\sqrt{5}}{2}(a+b)=\frac{\sqrt{5}}{2}.2(a+b+c)=\sqrt{5}(a+b+c)[/tex]
Có: [tex]\sqrt{a}+\sqrt{b}+\sqrt{c}=1\rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2=1\leq 3(a+b+c)\rightarrow a+b+c\geq \frac{1}{3}[/tex]
Vậy
[tex]\sqrt{5}(a+b+c)\geq \frac{\sqrt{5}}{3}[/tex]
Dấu = xảy ra khi a=b=c=1/9