Đặt \(\left(\sqrt{5a+4};\sqrt{5b+4};\sqrt{5c+4}\right)=\left(x;y;z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2=17\\2\le x;y;z\le3\end{matrix}\right.\)
\(P=x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=\sqrt{51}\)
\(P_{max}=\sqrt{51}\) khi \(a=b=c=\frac{1}{3}\)
\(2\le x\le3\Rightarrow\left(x-2\right)\left(x-3\right)\le0\Rightarrow x\ge\frac{x^2+6}{5}\)
Tương tự: \(y\ge\frac{y^2+6}{5}\) ; \(z\ge\frac{z^2+6}{5}\)
Cộng vế với vế: \(P\ge\frac{x^2+y^2+z^2+18}{5}=7\)
\(P_{min}=7\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị