\(P\le\sqrt{3\left(\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}\right)}=\sqrt{3\left(a+b+c\right)}\le\sqrt{3\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt[4]{27}\)
\(P_{max}=\sqrt[4]{27}\) khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Do \(\left\{{}\begin{matrix}0\le a;b;c\\a^2+b^2+c^2\le1\end{matrix}\right.\) \(\Rightarrow0\le a;b;c\le1\)
\(\Rightarrow\left\{{}\begin{matrix}a\left(a-1\right)\le0\\b\left(b-1\right)\le0\\c\left(c-1\right)\le0\end{matrix}\right.\) \(\Rightarrow a+b+c\ge a^2+b^2+c^2\)
Ta có:
\(P^2=a+b+c+2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{4}}+2\sqrt{\frac{\left(b+c\right)\left(c+a\right)}{4}}+2\sqrt{\frac{\left(a+b\right)\left(c+a\right)}{4}}\)
\(P^2=a+b+c+\sqrt{a^2+ab+bc+ca}+\sqrt{b^2+ab+bc+ca}+\sqrt{c^2+ab+bc+ca}\)
\(P^2\ge a+b+c+\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}=2\left(a+b+c\right)\ge2\left(a^2+b^2+c^2\right)=2\)
\(\Rightarrow P\ge\sqrt{2}\)
\(P_{min}=\sqrt{2}\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị
