Bài này có max chứ ko có min, chắc bạn ghi nhầm đề?
\(a^2+b^2-ab+b^2+1+b^2\ge2ab-ab+2b+b^2=b\left(a+b+2\right)\)
\(\Rightarrow P\le\sum\frac{1}{\sqrt{b\left(a+b+2\right)}}=\sum\frac{2}{\sqrt{4b\left(a+b+2\right)}}\le\sum\left(\frac{1}{4b}+\frac{1}{a+b+1+1}\right)\)
\(P\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\sum\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{1}+\frac{1}{1}\right)\le\frac{3}{4}+\frac{1}{16}\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}+6\right)\le\frac{3}{2}\)
\(p_{max}=\frac{3}{2}\) khi \(a=b=c=1\)