\(\frac{3}{4}-P=\frac{1}{4}\Sigma_{cyc}\frac{\left(a-b\right)^2}{\left(2a+b+c\right)\left(2b+c+a\right)}\ge0\)
Vậy \(P\le\frac{3}{4}\)
Cách 2: \(P=\Sigma_{cyc}\frac{a}{2a+b+c}\le\Sigma_{cyc}\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)=\frac{3}{4}\)