Ta có:\(P=\sum\frac{a}{\sqrt{1-a}}\)
\(P=\sum\frac{a}{\sqrt{b+c}}\)
\(P\ge\sum\frac{\sqrt{\frac{8}{3}}a}{b+c+\frac{2}{3}}=\sum\sqrt{\frac{8}{3}}\frac{a^2}{ab+ac+\frac{2}{3}a}\)
\(P\ge\sqrt{\frac{8}{3}}\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+\frac{2}{3}\left(a+b+c\right)}\left(cauchy-sch\text{w}arz\right)\)
Mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)
\(\Rightarrow P\ge\sqrt{\frac{8}{3}}\frac{1}{\frac{2}{3}+\frac{2}{3}}=\sqrt{\frac{8}{3}}.\frac{3}{4}=\sqrt{\frac{3}{2}}\)
"="<=>a=b=c=1/3